LeetCode_27

题目描述：

Given an array nums and a value val, remove all instances of that value in-placeand return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn‘t matter what you leave beyond the new length.

思路：

定义区间 [0,k)范围为无val的数列，遍历nums

代码如下：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        if(nums.size() == 0) return 0;
        int k = 0;
        for(int i = 0; i < nums.size(); ++i)
        {
            if(nums[i] != val)
                nums[k++] = nums[i];
        }
        return k;
    }
};

LeetCode_27

原文：https://www.cnblogs.com/harchar/p/10502029.html