[LeetCode] 2297. Jump Game VIII
You are given a 0-indexed integer array nums
of length n
. You are initially standing at index 0
. You can jump from index i
to index j
where i < j
if:
nums[i] <= nums[j]
andnums[k] < nums[i]
for all indexesk
in the rangei < k < j
, ornums[i] > nums[j]
andnums[k] >= nums[i]
for all indexesk
in the rangei < k < j
.
You are also given an integer array costs
of length n
where costs[i]
denotes the cost of jumping to index i
.
Return the minimum cost to jump to the index n - 1
.
Example 1:
Input: nums = [3,2,4,4,1], costs = [3,7,6,4,2] Output: 8 Explanation: You start at index 0. - Jump to index 2 with a cost of costs[2] = 6. - Jump to index 4 with a cost of costs[4] = 2. The total cost is 8. It can be proven that 8 is the minimum cost needed. Two other possible paths are from index 0 -> 1 -> 4 and index 0 -> 2 -> 3 -> 4. These have a total cost of 9 and 12, respectively.
Example 2:
Input: nums = [0,1,2], costs = [1,1,1] Output: 2 Explanation: Start at index 0. - Jump to index 1 with a cost of costs[1] = 1. - Jump to index 2 with a cost of costs[2] = 1. The total cost is 2. Note that you cannot jump directly from index 0 to index 2 because nums[0] <= nums[1].
Constraints:
n == nums.length == costs.length
1 <= n <= 105
0 <= nums[i], costs[i] <= 105
跳跃游戏 VIII。
给定一个长度为 n 的下标从 0 开始的整数数组 nums。初始位置为下标 0。当 i < j 时,你可以从下标 i 跳转到下标 j:
对于在 i < k < j 范围内的所有下标 k 有 nums[i] <= nums[j] 和 nums[k] < nums[i] , 或者
对于在 i < k < j 范围内的所有下标 k 有 nums[i] > nums[j] 和 nums[k] >= nums[i] 。
你还得到了一个长度为 n 的整数数组 costs,其中 costs[i] 表示跳转到下标 i 的代价。返回跳转到下标 n - 1 的最小代价。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/jump-game-viii
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思路是单调栈 + 动态规划。注意题目规定的从下表 i 跳到下标 j 的条件有两个,
- 对于在 i < k < j 范围内的所有下标 k 有 nums[i] <= nums[j] 和 nums[k] < nums[i]
- 对于在 i < k < j 范围内的所有下标 k 有 nums[i] > nums[j] 和 nums[k] >= nums[i]
我们用中文翻译一下这两个条件,
- 条件一,j 一定在 i 的右侧,且是 i 的右侧遇到的第一个 >= nums[i] 的元素,因为中间的 nums[k] 统统小于 nums[i]
- 条件二,j 一定在 i 的右侧,且是 i 的右侧遇到的第一个 < nums[i] 的元素,因为中间的 nums[k] 统统大于等于 nums[i]
我们发现其实这两个条件是在逻辑上是互补的,所以无论 nums[j] 与 nums[i] 谁大谁小,都会被处理到。所以这里我们用两个 stack 分别处理条件一和条件二,这样我们就能确保从 0 开始可以跳到数组内的任意一个位置,无非是 cost 不同而已。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public long minCost(int[] nums, int[] costs) { 3 Deque<Integer> stack1 = new ArrayDeque<>(); 4 Deque<Integer> stack2 = new ArrayDeque<>(); 5 int n = nums.length; 6 long[] dp = new long[n]; 7 Arrays.fill(dp, Long.MAX_VALUE); 8 dp[0] = 0; 9 for (int j = 0; j < n; j++) { 10 while (!stack1.isEmpty() && nums[stack1.peek()] <= nums[j]) { 11 int i = stack1.pop(); 12 dp[j] = Math.min(dp[j], dp[i] + costs[j]); 13 } 14 stack1.push(j); 15 16 while (!stack2.isEmpty() && nums[stack2.peek()] > nums[j]) { 17 int i = stack2.pop(); 18 dp[j] = Math.min(dp[j], dp[i] + costs[j]); 19 } 20 stack2.push(j); 21 } 22 return dp[n - 1]; 23 } 24 }