Replace Elements with Greatest Element on Right Side
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
思路一:从数组后面开始遍历,用一个下标记录当前遍历的最大值,依次对比数组的每个位置即可
public int[] replaceElements(int[] arr) {
int[] max = new int[arr.length];
max[arr.length - 1] = -1;
int x = arr[arr.length - 1];
for (int i = arr.length - 2; i >= 0; i--) {
if (arr[i] > x) {
max[i] = x;
x = arr[i];
} else {
max[i] = x;
}
}
return max;
}
标签:index,arr,right,--,element,int,1299,easy,leetcode
From: https://www.cnblogs.com/iyiluo/p/16928249.html