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接外包项目_图司机在线海报制作_seo新手教程_sem扫描电镜

2025/3/1 20:25:50 来源:https://blog.csdn.net/weixin_46841376/article/details/145668697  浏览:    关键词:接外包项目_图司机在线海报制作_seo新手教程_sem扫描电镜
接外包项目_图司机在线海报制作_seo新手教程_sem扫描电镜

文章目录

    • 一、题目
    • 二、题解

一、题目

Given an array arr, replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1.

After doing so, return the array.

Example 1:

Input: arr = [17,18,5,4,6,1]
Output: [18,6,6,6,1,-1]
Explanation:

  • index 0 --> the greatest element to the right of index 0 is index 1 (18).
  • index 1 --> the greatest element to the right of index 1 is index 4 (6).
  • index 2 --> the greatest element to the right of index 2 is index 4 (6).
  • index 3 --> the greatest element to the right of index 3 is index 4 (6).
  • index 4 --> the greatest element to the right of index 4 is index 5 (1).
  • index 5 --> there are no elements to the right of index 5, so we put -1.
    Example 2:

Input: arr = [400]
Output: [-1]
Explanation: There are no elements to the right of index 0.

Constraints:

1 <= arr.length <= 104
1 <= arr[i] <= 105

二、题解

逆序遍历

class Solution {
public:vector<int> replaceElements(vector<int>& arr) {int n = arr.size();int maxV = arr[n-1];arr[n-1] = -1;for(int i = n - 2;i >= 0;i--){int tmp = arr[i];arr[i] = maxV;if(tmp > maxV) maxV = tmp;}return arr;}
};

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