LeetCode刷题第八九十周

动态规划

如果某一问题有很多重叠子问题，使用动态规划是最有效的

解题步骤：

背包问题：01背包，完全背包，多重背包

01背包：
统一使用一维数组来进行遍历

 public static void main(String[] args) {
        int[] weight = {1, 3, 4};
        int[] value = {15, 20, 30};
        int bagWight = 4;
        testWeightBagProblem(weight, value, bagWight);
    }

    public static void testWeightBagProblem(int[] weight, int[] value, int bagWeight){
        int wLen = weight.length;
        //定义dp数组：dp[j]表示背包容量为j时，能获得的最大价值
        int[] dp = new int[bagWeight + 1];
        //遍历顺序：先遍历物品，再遍历背包容量
        for (int i = 0; i < wLen; i++){
            for (int j = bagWeight; j >= weight[i]; j--){
                dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]);
                //对于组合问题的递推公式为：dp[j] += dp[j - nums[i]]，需要初始化dp[0]=1;
            }
        }
        //打印dp数组
        for (int j = 0; j <= bagWeight; j++){
            System.out.print(dp[j] + " ");
        }
    }

完全背包：

01背包和完全背包唯一不同就是体现在遍历顺序上,完全背包的物品是可以添加多次的，所以要从小到大去遍历
// 先遍历物品，再遍历背包
for(int i = 0; i < weight.size(); i++) { // 遍历物品
    for(int j = weight[i]; j <= bagWeight ; j++) { // 遍历背包容量
        dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        //对于组合问题的递推公式为：dp[j] += dp[j - nums[i]]，需要初始化dp[0]=1;
    }
}
//对于排列问题，先遍历背包，再遍历物品
dp[0] = 1;
for (int j = 0; j <= amount; j++) { // 遍历背包容量
    for (int i = 0; i < coins.size(); i++) { // 遍历物品
        if (j - coins[i] >= 0) dp[j] += dp[j - coins[i]];
    }
}

509、斐波那契数

class Solution {
    public int fib(int n) {
        // if(n==1){
        //     return 1;
        // }
        // if(n==0){
        //     return 0;
        // }
        // return fib(n-1)+fib(n-2);
        if(n==1){
            return 1;
        }
        if(n==0){
            return 0;
        }
        int[] dp = new int[n+1];
        dp[0] = 0;
        dp[1] = 1;
        for(int i = 2; i < n+1; i++){
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
}

70、爬楼梯

class Solution {
    public int climbStairs(int n) {
        if(n==1) return n;
        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;
        for(int i = 3; i <= n; i++){
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}

746、使用最小花费爬楼梯

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int[] dp = new int[cost.length + 1];
        dp[0] = 0;
        dp[1] = 0;
        for(int i = 2; i <= cost.length; i++){
            dp[i] = Math.min(dp[i - 1] + cost[i - 1],dp[i - 2] + cost[i - 2]);
        }
        return dp[cost.length];
    }
}

62、 不同路径

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        dp[0][0] = 1;
        for(int i = 1; i < m; i++){
            dp[i][0] = 1;
        }
        for(int j = 1; j < n; j++){
            dp[0][j] = 1;
        }
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];

    }
}

63、不同路径 II

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        if(obstacleGrid[0][0] == 1){
            return 0;
        }else{
            dp[0][0] = 1;
        }
        for(int i = 1; i < m; i++){
            if(obstacleGrid[i][0] == 1){
                break;
            }else{
                dp[i][0] = 1;
            }
        }
        for(int j = 1; j < n; j++){
            if(obstacleGrid[0][j] == 1){
                break;
            }else{
                dp[0][j] = 1;
            }
        }
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                if(obstacleGrid[i][j] == 1){
                    continue;
                }else{
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                } 
            }
        }
        return dp[m - 1][n - 1];
    }
}

343、整数拆分

class Solution {
    public int integerBreak(int n) {
        // 方法一：贪心
        // if(n == 2){
        //     return 1;
        // }else if(n == 3){
        //     return 2;
        // }
        // int result = 1;
        // while(n > 0){
        //     if(n > 4){
        //         result *= 3;
        //         n = n - 3;
        //     }else if(n == 4){
        //         result *= 4;
        //         break;
        //     }else if(n == 3 || n == 2){
        //         result *= n;
        //         break;
        //     }
        // }
        // return result;
        // 方法二：动态规划
        int[] dp = new int[n];
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2; i < n; i++){
            int left = 0;
            int right = i - 1;
            while(left <= right){
                dp[i] = Math.max(dp[i],Math.max(left+1,dp[left])*Math.max(right+1,dp[right]));
                left++;
                right--;
            }
        }
        return dp[n-1];
    }
}

96、不同的二叉搜索树

class Solution {
    public int numTrees(int n) {
        if(n==1){
            return 1;
        }
        int[] dp = new int[n+1];
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        for(int i = 3; i <= n; i++){
            int temp = i-1;
            while(temp>=0){
                dp[i] += dp[temp]*dp[i-temp-1];
                temp--;
            }
        }
        return dp[n];
    }
}

416、分割等和子集

class Solution {
    public boolean canPartition(int[] nums) {
        int sum = 0;
        for(int i : nums){
            sum += i;
        }
        if(sum%2==1){
            return false;
        }
        sum=sum/2;
        int[] dp = new int[sum+1];
        for(int i = 0; i < nums.length; i++){
            for(int j = sum; j >= nums[i]; j--){
                dp[j] = Math.max(dp[j],dp[j-nums[i]]+nums[i]);
                if(dp[j]==sum){
                    return true;
                }
            }
        }
        return false;
    }
}

1049、最后一块石头的重量 II

class Solution {
    public int lastStoneWeightII(int[] stones) {
        int sum = 0;
        for(int i : stones){
            sum += i;
        }
        int target = sum/2;
        int[] dp = new int[target+1];
        for(int i = 0; i < stones.length; i++){
            for(int j = target; j >= stones[i]; j--){
                dp[j] = Math.max(dp[j],dp[j-stones[i]]+stones[i]);
            }
        }
        return sum-2*dp[target];
    }
}

494、目标和

class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int sum = 0;
        for(int i:nums){
            sum += i;
        }
        if(Math.abs(target) > sum) return 0;
        if((sum + target) % 2 == 1) return 0;
        int bageSize = (sum + target) / 2;
        int[] dp = new int[bageSize + 1];
        dp[0] = 1;
        for(int i = 0; i < nums.length; i++){
            for(int j = bageSize; j >= nums[i]; j--){
                //在求装满背包有几种方法的情况下，递推公式一般为
                dp[j] += dp[j - nums[i]];//组合问题
            }
        }
        return dp[bageSize];
    }
}

474、一和零

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        for(String str : strs){
            int zeroNum = 0;
            int oneNum = 0;
            for(char c: str.toCharArray()){
                if(c == '0'){
                    zeroNum++;
                }else{
                    oneNum++;
                } 
            }
            for(int i = m; i >= zeroNum; i--){
                for(int j = n; j >= oneNum; j--){
                    dp[i][j] = Math.max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1);
                }
            }
        }
        return dp[m][n];
    }
}

518、 零钱兑换 II

class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for(int i = 0; i < coins.length; i++){
            for(int j = coins[i]; j <= amount; j++){
                dp[j] += dp[j - coins[i]];
            }
        }
        return dp[amount];
    }
}

377、组合总和 Ⅳ

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target + 1];
        dp[0] = 1;
        for(int i = 1; i <= target; i++){
            for(int j = 0; j < nums.length; j++){
                if(i >= nums[j]){
                    dp[i] += dp[i - nums[j]];
                } 
            }
        }
        return dp[target];
    }
}

322、零钱兑换

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp,Integer.MAX_VALUE);
        dp[0] = 0;
        for(int i = 0; i < coins.length; i++){
            for(int j = coins[i]; j <= amount; j++){
                if(dp[j-coins[i]]!=Integer.MAX_VALUE)
                dp[j] = Math.min(dp[j],dp[j - coins[i]] + 1);
            }
        }
        return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
    }
}

279、完全平方数

class Solution {
    public int numSquares(int n) {
        int k = (int)Math.pow(n,0.5);
        int[] nums = new int[k];
        for(int i = 0; i < nums.length; i++){
            nums[i] = (i+1)*(i+1);
        }
        int[] dp = new int[n+1];
        Arrays.fill(dp,Integer.MAX_VALUE);
        dp[0] = 0;
        for(int i = 0; i < nums.length; i++){
            for(int j = nums[i]; j <= n; j++){
                if(dp[j - nums[i]]!=Integer.MAX_VALUE)
                dp[j] = Math.min(dp[j],dp[j - nums[i]] + 1);
            }
        }
        return dp[n];
    }
}

139、单词拆分

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for(int j = 0; j <= s.length(); j++){
            for(String str : wordDict){
                int len = str.length();
                if(j>=len&&s.substring(j-len,j).equals(str)&&dp[j-len]){
                    dp[j] = true;
                }
            }
        }
        return dp[s.length()];
    }
}

198、打家劫舍

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 1) return nums[0];
        int[] dp = new int[nums.length];
        dp[0] = nums[0];
        dp[1] = Math.max(dp[0], nums[1]);
        for(int i = 2; i < nums.length; i++){
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        return dp[nums.length - 1];
    }
}

213、打家劫舍 II

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 1) return nums[0];
        return Math.max(robAction(nums, 0, nums.length - 2), robAction(nums, 1, nums.length - 1));
    }
    public int robAction(int[] nums, int start, int end){
        if(start == end) return nums[start];
        int[] dp = new int[nums.length];
        dp[start] = nums[start];
        dp[start + 1] = Math.max(dp[start], nums[start + 1]);
        for(int i = start + 2; i <= end; i++){
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        return dp[end];
    }
}

337、打家劫舍 III

class Solution {
    public int rob(TreeNode root) {
// 后序遍历
        int[] res = robAction(root);
        return Math.max(res[0], res[1]);
    }
    public int[] robAction(TreeNode root){
        int[] res = new int[2];
        if(root == null) return res;
        int[] left = robAction(root.left);
        int[] right = robAction(root.right);
        res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        res[1] = root.val + left[0] + right[0];
        return res;
    }
}

121、买卖股票的最佳时机

class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;
        if(len == 1) return 0;
        int[][] dp = new int[len][2];
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for(int i = 1; i < len; i++){
            dp[i][0] = Math.max(dp[i - 1][0], -prices[i]);
            dp[i][1] = Math.max(dp[i - 1][0] + prices[i], dp[i - 1][1]);
        }
        return dp[len - 1][1];
    }
}

122、买卖股票的最佳时机 II

class Solution {
    public int maxProfit(int[] prices) {
        // 贪心
        // int max = 0;
        // for(int i = 1; i < prices.length; i++){
        //     if(prices[i] > prices[i - 1]){
        //         max += prices[i] - prices[i - 1];
        //     }
        // }
        // return max;
        // 动态规划
        int len = prices.length;
        if(len == 1) return 0;
        int[][] dp = new int[len][2];
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for(int i = 1; i < len; i++){
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] -prices[i]);
            dp[i][1] = Math.max(dp[i - 1][0] + prices[i], dp[i - 1][1]);
        }
        return dp[len - 1][1];
    }
}

123、买卖股票的最佳时机 III

class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;
        int[][] dp = new int[len][5];
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
        for(int i = 1; i < len; i++){
            dp[i][1] = Math.max(-prices[i], dp[i - 1][1]);
            dp[i][2] = Math.max(dp[i - 1][1] + prices[i], dp[i - 1][2]);
            dp[i][3] = Math.max(dp[i - 1][2] - prices[i], dp[i - 1][3]);
            dp[i][4] = Math.max(dp[i - 1][3] + prices[i], dp[i - 1][4]);
        }
        return dp[len - 1][4];
    }
}

188、买卖股票的最佳时机 IV

class Solution {
    public int maxProfit(int k, int[] prices) {
        int len = prices.length;
        int[][] dp = new int[len][1 + 2 * k];
        for(int i = 1; i < 1 + 2 * k; i = i + 2){
            dp[0][i] = -prices[0];
        }
        for(int i = 1; i < len; i++){
            for(int j = 1; j < 1 + 2 * k; j++){
                if(j == 1){
                    dp[i][j] = Math.max(-prices[i], dp[i - 1][1]);
                }else if(j%2==0){
                    dp[i][j] = Math.max(dp[i - 1][j - 1] + prices[i], dp[i - 1][j]);
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j - 1] - prices[i], dp[i - 1][j]);
                }
            }
        }
        return dp[len - 1][2 * k];
    }
}

309、最佳买卖股票时机含冷冻期

class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;
        if(len == 1) return 0;
        int[][] dp = new int[len][4];
        dp[0][0] = -prices[0];
        for(int i = 1; i < len; i++){
            dp[i][0] = Math.max(dp[i - 1][0], Math.max(dp[i - 1][1], dp[i - 1][3]) - prices[i]);
            dp[i][1] = Math.max(dp[i - 1][3], dp[i - 1][1]);
            dp[i][2] = dp[i - 1][0] + prices[i];
            dp[i][3] = dp[i - 1][2];
        }
        return Math.max(dp[len - 1][1], Math.max(dp[len - 1][2], dp[len - 1][3]));
    }
}

714、买卖股票的最佳时机含手续费

class Solution {
    public int maxProfit(int[] prices, int fee) {
        // 贪心
        // if(prices.length==1){
        //     return 0;
        // }
        // int index = prices[0]+fee;
        // int sum=0;
        // for(int i = 1; i < prices.length; i++){
        //     if(prices[i]>index){
        //         sum+=prices[i]-index;
        //         index=prices[i];
        //     }else if(prices[i]+fee<index){
        //         index=prices[i]+fee;
        //     }
        // }
        // return sum;
        // 动态规划
        int len = prices.length;
        if(len == 1) return 0;
        int[][] dp = new int[len][2];
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for(int i = 1; i < len; i++){
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] -prices[i]);
            dp[i][1] = Math.max(dp[i - 1][0] + prices[i] - fee, dp[i - 1][1]);
        }
        return Math.max(dp[len - 1][1],dp[len - 1][0]);
    }
}

300、最长递增子序列

class Solution {
    public int lengthOfLIS(int[] nums) {
        int[] dp = new int[nums.length];
        Arrays.fill(dp,1);
        int result = 1;
        for(int i = 1; i < nums.length; i++){
            for(int j = 0; j < i; j++){
                if(nums[i] > nums[j]) dp[i] = Math.max(dp[i], dp[j] + 1);
            }
            if(dp[i] > result) result = dp[i];
        }
        return result;
    }
}

674、最长连续递增序列

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int[] dp = new int[nums.length];
        Arrays.fill(dp,1);
        int result = 1;
        for(int i = 1; i < nums.length; i++){
            if(nums[i]>nums[i-1]){
                dp[i] = Math.max(dp[i-1]+1,dp[i]);
                result = result > dp[i] ? result : dp[i];
            }
        }
        return result;
    }
}

718、最长重复子数组

class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        int result = 0;
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
            if(dp[i][j] > result) result = dp[i][j];
            }
        }
        return result;
    }
}

1143、最长公共子序列

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        char[] nums1 = text1.toCharArray();
        char[] nums2 = text2.toCharArray();
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        int result = 0;
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            if(dp[i][j] > result) result = dp[i][j];
            }
        }
        return result;
    }
}

1035、不相交的线

class Solution {
    public int maxUncrossedLines(int[] nums1, int[] nums2) {
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        int result = 0;
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            if(dp[i][j] > result) result = dp[i][j];
            }
        }
        return result;
    }
}

53、最大子数组和

class Solution {
    public int maxSubArray(int[] nums) {
        // 贪心
        // if(nums.length == 1){
        //     return nums[0];
        // }
        // int count = 0;
        // int sum = Integer.MIN_VALUE;
        // for(int i = 0; i < nums.length; i++){
        //     count += nums[i];
        //     sum = Math.max(sum, count);
        //     if(count <= 0){
        //         count = 0;
        //     }
        // }
        // return sum;
        // 动态规划
        int[] dp = new int[nums.length];
        dp[0] = nums[0];
        int result = nums[0];
        for(int i = 1; i < nums.length; i++){
            dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]);
            result = dp[i] > result ? dp[i] : result;
        }
        return result;
    }
}

392、 判断子序列

class Solution {
    public boolean isSubsequence(String s, String t) {
        char[] nums1 = s.toCharArray();
        char[] nums2 = t.toCharArray();
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        int result = 0;
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            if(dp[i][j] > result) result = dp[i][j];
            }
        }
        return result == s.length();
    }
}

115、不同的子序列

class Solution {
    public int numDistinct(String s, String t) {
        char[] nums1 = s.toCharArray();
        char[] nums2 = t.toCharArray();
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        for(int i = 0; i < nums1.length; i++){
            dp[i][0] = 1;
        }
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }else{
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[nums1.length][nums2.length];
    }
}

583、两个字符串的删除操作

class Solution {
    public int minDistance(String word1, String word2) {
        char[] nums1 = word1.toCharArray();
        char[] nums2 = word2.toCharArray();
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        int result = 0;
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            if(dp[i][j] > result) result = dp[i][j];
            }
        }
        return word1.length() + word2.length() - 2 * result;
    }
}

72、编辑距离

class Solution {
    public int minDistance(String word1, String word2) {
        char[] nums1 = word1.toCharArray();
        char[] nums2 = word2.toCharArray();
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        int result = 0;
        for(int i = 0; i <= nums1.length; i++){
            dp[i][0] = i;
        }
        for(int i = 0; i <= nums2.length; i++){
            dp[0][i] = i;
        }
        for(int i = 1; i <= nums1.length; i++){
            for(int j = 1; j <= nums2.length; j++){
                if(nums1[i - 1] == nums2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1];
                }else{
                    dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;;
                }
            }
        }
        return dp[nums1.length][nums2.length];
    }
}

647、回文子串

class Solution {
    public int countSubstrings(String s) {
        boolean[][] dp = new boolean[s.length()][s.length()];
        int count = 0;
        for(int i = dp.length-1; i >= 0; i--){
            for(int j = i; j < dp.length; j++){
                if(s.charAt(i)==s.charAt(j)){
                    if(j-i<=1){
                        count++;
                        dp[i][j] = true;
                    }else if(dp[i+1][j-1]){
                        count++;
                        dp[i][j] = true;
                    }
                }
                
            }
        }
        return count;
    }
}

516、最长回文子序列

class Solution {
    public int longestPalindromeSubseq(String s) {
        int[][] dp = new int[s.length() + 1][s.length() + 1];
        int result = 1;
        for(int i = s.length() - 1; i >= 0; i--){
            dp[i][i] = 1;
            for(int j = i + 1; j < s.length(); j++){
                if(s.charAt(i) == s.charAt(j)){
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                }else{
                    dp[i][j] = Math.max(dp[i][j], Math.max(dp[i + 1][j], dp[i][j - 1]));
                }
                result = result > dp[i][j] ? result : dp[i][j];
            }
        }
        return result;
    }
}

739、每日温度

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int lens = temperatures.length;
        int[] res = new int[lens];
        Stack<Integer> stack = new Stack<>();
        stack.push(0);
        for(int i = 1; i < lens; i++){
            if(temperatures[i] <= temperatures[stack.peek()]){
                stack.push(i);
            }else{
                while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]){
                    int temp = stack.pop();
                    res[temp] = i - temp;
                }
                stack.push(i);
            }
        }
        return res;
    }
}

496、下一个更大元素 I

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int lens = nums1.length;
        int[] res = new int[lens];
        Arrays.fill(res,-1);
        Stack<Integer> stack = new Stack<>();
        HashMap<Integer, Integer> hashMap = new HashMap<>();
        for (int i = 0 ; i< nums1.length ; i++){
            hashMap.put(nums1[i],i);
        }
        stack.push(0);
        for(int i = 1; i < nums2.length; i++){
            if(nums2[i] <= nums2[stack.peek()]){
                stack.push(i);
            }else{
                while(!stack.isEmpty() && nums2[i] > nums2[stack.peek()]){
                    if (hashMap.containsKey(nums2[stack.peek()])){
                        Integer index = hashMap.get(nums2[stack.peek()]);
                        res[index] = nums2[i];
                    }
                    stack.pop();
                }
                stack.push(i);
            }
        }
        return res;
    }
}

503、下一个更大元素 II

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int lens = nums.length;
        int[] res = new int[lens];
        Arrays.fill(res,-1);
        Stack<Integer> stack = new Stack<>();
        stack.push(0);
        for(int i = 1; i < 2*lens; i++){
            if(nums[i%lens] <= nums[stack.peek()]){
                stack.push(i%lens);
            }else{
                while(!stack.isEmpty() && nums[i%lens] > nums[stack.peek()]){
                    int temp = stack.pop();
                    res[temp] = nums[i%lens];
                }
                stack.push(i%lens);
            }
        }
        return res;
    }
}

[42]
[84]

搜索插入位置

35

class Solution {
    public int searchInsert(int[] nums, int target) {
        // 有序数组，考虑用二分查找
        int left = 0;
        int right = nums.length - 1;
        int mid = (left + right) >> 1;
        if(target < nums[left]){
            return left;
        }
        if(target > nums[right]){
            return right + 1;
        }
        while(left <= right){
            if(target == nums[mid]){
                return mid;
            }else if(target < nums[mid]){
                right = mid -1;
            }else{
                left = mid + 1;
            }
            mid = (left + right) >> 1;
        }
        return left;//找不到，返回需要插入的位置
    }
}

在排序数组中查找元素的第一个和最后一个位置

34

class Solution {
    public int[] searchRange(int[] nums, int target) {
        // 非递减说明是升序的，但可以有重复元素
        int[] arr = {-1, -1};
        if(nums.length == 0){
            return arr;
        }
        int left = 0;
        int right = nums.length - 1;
        int mid = (left + right) >> 1;
        if(target < nums[left] || target > nums[right]){
            return arr;//边界值
        }
        int leftPoint;//目标数组的开始位置
        int rightPoint;//目标数组的结束位置
        while(left <= right){
            if(target == nums[mid]){
                leftPoint = mid;
                rightPoint = mid;
                while(leftPoint >= 0 && target == nums[leftPoint]){
                    arr[0] = leftPoint;
                    leftPoint--;//向左寻找重复元素
                }
                while(rightPoint <= (nums.length - 1) && target == nums[rightPoint]){
                    arr[1] = rightPoint;
                    rightPoint++;//向右寻找重复元素
                }
                return arr;//返回找到的目标值的位置
            }else if(target < nums[mid]){
                right = mid - 1;
            }else{
                left = mid + 1;
            }
            mid = (left + right) >> 1;
        }
        return arr;//没有找到
    }
}