27. Remove Element

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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.

Hint:

Try two pointers.
Did you use the property of “the order of elements can be changed”?
What happens when the elements to remove are rare?

思路：这道题目与 26.Remove Duplicates from Sorted Array
思路和做法都十分相似。方法是仍然使用i 和index两个从零开始的下标，使用for循环来遍历nums数组，每当nums[i] != val, 就把这个nums[i] 赋给nums[index]。同样，还是需要注意当nums数组为空的情况。

代码如下：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
    if(nums.empty()) return 0;

    int index = 0;
    for(int i = 0; i < nums.size(); i++){
        if(nums[i] != val)
            nums[index++] = nums[i];
    }
    return index; //因为上一行是index++,所以返回的index已经+1，就是新数组长度
    }
};

相关题目：
26.Remove Duplicates from Sorted Array