--- comments: true difficulty: 中等 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0962.Maximum%20Width%20Ramp/README.md tags: - 栈 - 数组 - 单调栈 --- # [962. 最大宽度坡](https://leetcode.cn/problems/maximum-width-ramp) [English Version](/solution/0900-0999/0962.Maximum%20Width%20Ramp/README_EN.md) ## 题目描述

给定一个整数数组 A是元组 (i, j),其中  i < j 且 A[i] <= A[j]。这样的坡的宽度为 j - i

找出 A 中的坡的最大宽度,如果不存在,返回 0 。

 

示例 1:

输入:[6,0,8,2,1,5]
输出:4
解释:
最大宽度的坡为 (i, j) = (1, 5): A[1] = 0 且 A[5] = 5.

示例 2:

输入:[9,8,1,0,1,9,4,0,4,1]
输出:7
解释:
最大宽度的坡为 (i, j) = (2, 9): A[2] = 1 且 A[9] = 1.

 

提示:

  1. 2 <= A.length <= 50000
  2. 0 <= A[i] <= 50000

 

## 解法 ### 方法一:单调栈 根据题意,我们可以发现,所有可能的 $nums[i]$ 所构成的子序列一定是单调递减的。为什么呢?我们不妨用反证法证明一下。 假设存在 $i_1 #### Python3 ```python class Solution: def maxWidthRamp(self, nums: List[int]) -> int: stk = [] for i, v in enumerate(nums): if not stk or nums[stk[-1]] > v: stk.append(i) ans = 0 for i in range(len(nums) - 1, -1, -1): while stk and nums[stk[-1]] <= nums[i]: ans = max(ans, i - stk.pop()) if not stk: break return ans ``` #### Java ```java class Solution { public int maxWidthRamp(int[] nums) { int n = nums.length; Deque stk = new ArrayDeque<>(); for (int i = 0; i < n; ++i) { if (stk.isEmpty() || nums[stk.peek()] > nums[i]) { stk.push(i); } } int ans = 0; for (int i = n - 1; i >= 0; --i) { while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) { ans = Math.max(ans, i - stk.pop()); } if (stk.isEmpty()) { break; } } return ans; } } ``` #### C++ ```cpp class Solution { public: int maxWidthRamp(vector& nums) { int n = nums.size(); stack stk; for (int i = 0; i < n; ++i) { if (stk.empty() || nums[stk.top()] > nums[i]) stk.push(i); } int ans = 0; for (int i = n - 1; i; --i) { while (!stk.empty() && nums[stk.top()] <= nums[i]) { ans = max(ans, i - stk.top()); stk.pop(); } if (stk.empty()) break; } return ans; } }; ``` #### Go ```go func maxWidthRamp(nums []int) int { n := len(nums) stk := []int{} for i, v := range nums { if len(stk) == 0 || nums[stk[len(stk)-1]] > v { stk = append(stk, i) } } ans := 0 for i := n - 1; i >= 0; i-- { for len(stk) > 0 && nums[stk[len(stk)-1]] <= nums[i] { ans = max(ans, i-stk[len(stk)-1]) stk = stk[:len(stk)-1] } if len(stk) == 0 { break } } return ans } ``` #### TypeScript ```ts function maxWidthRamp(nums: number[]): number { let [ans, n] = [0, nums.length]; const stk: number[] = []; for (let i = 0; i < n - 1; i++) { if (stk.length === 0 || nums[stk.at(-1)!] > nums[i]) { stk.push(i); } } for (let i = n - 1; i >= 0; i--) { while (stk.length && nums[stk.at(-1)!] <= nums[i]) { ans = Math.max(ans, i - stk.pop()!); } if (stk.length === 0) break; } return ans; } ``` #### JavaScript ```js function maxWidthRamp(nums) { let [ans, n] = [0, nums.length]; const stk = []; for (let i = 0; i < n - 1; i++) { if (stk.length === 0 || nums[stk.at(-1)] > nums[i]) { stk.push(i); } } for (let i = n - 1; i >= 0; i--) { while (stk.length && nums[stk.at(-1)] <= nums[i]) { ans = Math.max(ans, i - stk.pop()); } if (stk.length === 0) break; } return ans; } ``` ### 方法二:排序 #### TypeScript ```ts function maxWidthRamp(nums: number[]): number { const idx = nums.map((x, i) => [x, i]).sort(([a], [b]) => a - b); let [ans, j] = [0, nums.length]; for (const [_, i] of idx) { ans = Math.max(ans, i - j); j = Math.min(j, i); } return ans; } ``` #### JavaScript ```js function maxWidthRamp(nums) { const idx = nums.map((x, i) => [x, i]).sort(([a], [b]) => a - b); let [ans, j] = [0, nums.length]; for (const [_, i] of idx) { ans = Math.max(ans, i - j); j = Math.min(j, i); } return ans; } ```