Description
Given an array and a value, remove all occurrences of that value in place and return the new length.
The order of elements can be changed, and the elements after the new length don‘t matter.
Example
Given an array [0,4,4,0,0,2,4,4]
, value=4
return 4
and front four elements of the array is [0,0,0,2]
解题:今天这个题目还是挺简单的,给定一个数组,再给定一个值,要求把数组中里存在该值的,都剔除掉。在原数组的基础上,定一个rear作为待插入位置的下标,从0开始。遍历数组,如果不是value,放在rear位置,如果是,则啥也不做,直到循环结束。最后返回rear值,即除去value的元素个数。代码如下:
1 public class Solution { 2 /* 3 * @param A: A list of integers 4 * @param elem: An integer 5 * @return: The new length after remove 6 */ 7 public int removeElement(int[] A, int elem) { 8 // write your code here 9 int rear = 0; 10 for(int i = 0; i < A.length; i++){ 11 if(A[i] != elem){ 12 A[rear++] = A[i]; 13 } 14 } 15 return rear; 16 } 17 }
原文地址:https://www.cnblogs.com/phdeblog/p/9141379.html
时间: 2024-11-12 08:54:36