[LeetCode] 1299. Replace Elements with Greatest Element on Right Side 将每个元素替换为右侧最大元素
作者：互联网
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 <= 1041 <= arr[i] <= 105
这道题给了一个数组 arr，说是让把每个数字更新为其右边的数字中最大的一个，最后一个数字变为 -1。既然是一道 Easy 的题目，就不用担心解法会太复杂，一般都是较短的行数就能搞定的。让求每个数字右边的数字中最大的一个，当然不能每次都遍历右边所有的数字来找最大值，虽说是 Easy 题目，但最好也别这么暴力地解，多少还是要给 OJ 一些尊重的。从左往右不好使的话，可以调个头，从右往左去更新，这样就简单的多了，最后一个数字直接更新为 -1，然后只要维护一个从右往左的当前最大值，每次用来更新右往左的对应位置即可，基本没什么难度，只要能想到换个方向更新，基本就迎刃而解了，参见代码如下：
class Solution {
public:
vector<int> replaceElements(vector<int>& arr) {
int n = arr.size(), curMax = INT_MIN;
vector<int> res(n, -1);
for (int i = n - 2; i >= 0; --i) {
curMax = max(curMax, arr[i + 1]);
res[i] = curMax;
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1299
类似题目：
Two Furthest Houses With Different Colors
参考资料：
https://leetcode.com/problems/replace-elements-with-greatest-element-on-right-side/
LeetCode All in One 题目讲解汇总(持续更新中...)
标签：index,arr,Elements,--,元素,element,right,greatest,Side 来源： https://www.cnblogs.com/grandyang/p/16284279.html