leetcode(27) - Remove Element

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.

int removeElement(int* nums, int numsSize, int val) {    int i=0, start=0;        while(start < numsSize ) {   // 首先找到第一个相等的位置          if(nums[start] == val)             break;                      start++;    }        if (start < numsSize-1) {   // 然后后面只要找到不相等的nums[i]，依次覆盖nums[start++]                i=start+1;                while (i < numsSize ) {            if(nums[i] != val)                nums[start++] = nums[i];                            i++;        }    }            return start;    // 返回处理后的数组的个数    }