代码随想录——单调栈
每日温度
题目 中等
什么时候用单调栈呢？
通常是一维数组，要寻找任一个元素的右边或者左边第一个比自己大或者小的元素的位置，此时我们就要想到可以用单调栈了。
class Solution {
public int[] dailyTemperatures(int[] temperatures) {
int lens = temperatures.length;
int[] res = new int[lens];
Deque<Integer> stack = new LinkedList<>();
for (int i = 0; i < lens; i++) {
while (!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
res[stack.peek()] = i - stack.peek();
stack.pop();
}
stack.push(i);
}
return res;
}
}下一个更大元素 I
题目 简单
class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums1.length; i++) {
map.put(nums1[i], i);
}
int[] res = new int[nums1.length];
Stack<Integer> stack = new Stack<>();
Arrays.fill(res, -1);
for (int i = 0; i < nums2.length; i++) {
while (!stack.isEmpty() && nums2[stack.peek()] < nums2[i]) {
int pre = nums2[stack.pop()];
if (map.containsKey(pre)) {
res[map.get(pre)] = nums2[i];
}
}
stack.push(i);
}
return res;
}
}下一个更大元素 II
题目 中等
本题我觉得我的解法更优一点。以下是代码随想录上的解法：
class Solution {
public int[] nextGreaterElements(int[] nums) {
//边界判断
if (nums == null || nums.length <= 1) {
return new int[]{-1};
}
int size = nums.length;
int[] result = new int[size];//存放结果
Arrays.fill(result, -1);//默认全部初始化为-1
Stack<Integer> st = new Stack<>();//栈中存放的是nums中的元素下标
for (int i = 0; i < 2 * size; i++) {
while (!st.empty() && nums[i % size] > nums[st.peek()]) {
result[st.peek()] = nums[i % size];//更新result
st.pop();//弹出栈顶
}
st.push(i % size);
}
return result;
}
}接雨水
题目 困难
认真去理解题解
方法一：动态规划
class Solution {
public int trap(int[] height) {
int n = height.length;
if (n == 0) {
return 0;
}
int[] leftMax = new int[n];
leftMax[0] = height[0];
for (int i = 1; i < n; ++i) {
leftMax[i] = Math.max(leftMax[i - 1], height[i]);
}
int[] rightMax = new int[n];
rightMax[n - 1] = height[n - 1];
for (int i = n - 2; i >= 0; --i) {
rightMax[i] = Math.max(rightMax[i + 1], height[i]);
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += Math.min(leftMax[i], rightMax[i]) - height[i];
}
return ans;
}
}
作者：力扣官方题解
链接：https://leetcode.cn/problems/trapping-rain-water/solutions/692342/jie-yu-shui-by-leetcode-solution-tuvc/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。方法二：单调栈
class Solution {
public int trap(int[] height) {
int ans = 0;
Deque<Integer> stack = new LinkedList<Integer>();
int n = height.length;
for (int i = 0; i < n; ++i) {
while (!stack.isEmpty() && height[i] > height[stack.peek()]) {
int top = stack.pop();
if (stack.isEmpty()) {
break;
}
int left = stack.peek();
int currWidth = i - left - 1;
int currHeight = Math.min(height[left], height[i]) - height[top];
ans += currWidth * currHeight;
}
stack.push(i);
}
return ans;
}
}
作者：力扣官方题解
链接：https://leetcode.cn/problems/trapping-rain-water/solutions/692342/jie-yu-shui-by-leetcode-solution-tuvc/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。方法三：双指针
class Solution {
public int trap(int[] height) {
int ans = 0;
int left = 0, right = height.length - 1;
int leftMax = 0, rightMax = 0;
while (left < right) {
leftMax = Math.max(leftMax, height[left]);
rightMax = Math.max(rightMax, height[right]);
if (height[left] < height[right]) {
ans += leftMax - height[left];
++left;
} else {
ans += rightMax - height[right];
--right;
}
}
return ans;
}
}
作者：力扣官方题解
链接：https://leetcode.cn/problems/trapping-rain-water/solutions/692342/jie-yu-shui-by-leetcode-solution-tuvc/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。柱状图中最大的矩形
题目 困难
详细看liweiwei大佬的题解
不加哨兵：
import java.util.ArrayDeque;
import java.util.Deque;
public class Solution {
public int largestRectangleArea(int[] heights) {
int len = heights.length;
if (len == 0) {
return 0;
}
if (len == 1) {
return heights[0];
}
int res = 0;
Deque<Integer> stack = new ArrayDeque<>(len);
for (int i = 0; i < len; i++) {
// 这个 while 很关键，因为有可能不止一个柱形的最大宽度可以被计算出来
while (!stack.isEmpty() && heights[i] < heights[stack.peekLast()]) {
int curHeight = heights[stack.pollLast()];
while (!stack.isEmpty() && heights[stack.peekLast()] == curHeight) {
stack.pollLast();
}
int curWidth;
if (stack.isEmpty()) {
curWidth = i;
} else {
curWidth = i - stack.peekLast() - 1;
}
// System.out.println("curIndex = " + curIndex + " " + curHeight * curWidth);
res = Math.max(res, curHeight * curWidth);
}
stack.addLast(i);
}
while (!stack.isEmpty()) {
int curHeight = heights[stack.pollLast()];
while (!stack.isEmpty() && heights[stack.peekLast()] == curHeight) {
stack.pollLast();
}
int curWidth;
if (stack.isEmpty()) {
curWidth = len;
} else {
curWidth = len - stack.peekLast() - 1;
}
res = Math.max(res, curHeight * curWidth);
}
return res;
}
}
作者：liweiwei1419
链接：https://leetcode.cn/problems/largest-rectangle-in-histogram/solutions/142012/bao-li-jie-fa-zhan-by-liweiwei1419/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。加哨兵：
import java.util.ArrayDeque;
import java.util.Deque;
public class Solution {
public int largestRectangleArea(int[] heights) {
int len = heights.length;
if (len == 0) {
return 0;
}
if (len == 1) {
return heights[0];
}
int res = 0;
int[] newHeights = new int[len + 2];
newHeights[0] = 0;
System.arraycopy(heights, 0, newHeights, 1, len);
newHeights[len + 1] = 0;
len += 2;
heights = newHeights;
Deque<Integer> stack = new ArrayDeque<>(len);
// 先放入哨兵，在循环里就不用做非空判断
stack.addLast(0);
for (int i = 1; i < len; i++) {
while (heights[i] < heights[stack.peekLast()]) {
int curHeight = heights[stack.pollLast()];
int curWidth = i - stack.peekLast() - 1;
res = Math.max(res, curHeight * curWidth);
}
stack.addLast(i);
}
return res;
}
}
作者：liweiwei1419
链接：https://leetcode.cn/problems/largest-rectangle-in-histogram/solutions/142012/bao-li-jie-fa-zhan-by-liweiwei1419/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
