Given an array and a value, remove all instances of that value in place and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
Example:
Given input array nums = [3,2,2,3], val = 3
Your function should return length = 2, with the first two elements of nums being 2.题意:给一个数组,再给你一个数val,从数组中把val删掉并返回剩余数组的长度
int removeElement(int* nums, int numsSize, int val) {
int i=0;
int j=0;
//这个就跟那个26题去重一个思想了
for(j=0;j<numsSize;j++){
if(nums[j]!=val){
nums[i++]=nums[j];
}
}
return i;
}PS:这个跟之前的26题一样,双指针,,,,,
转载于:https://blog.51cto.com/fulin0532/1868692
本文介绍了一种在原地删除数组中特定值的算法,并确保内存使用为常数。通过双指针技巧实现,该算法改变了原有数组元素的顺序,但能满足题目要求,即返回不含特定值的新数组长度。
7360

被折叠的 条评论
为什么被折叠?



