Description

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 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.

Example:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.

Solution

移除数组中和指定val相等的元素，返回剩余元素个数k，和26. Remove Duplicates from Sorted Array类似，保证前k个元素ok就行，后面n-k位置放什么不用管

int removeElement(vector<int>& nums, int val) {
    if (nums.empty()) {
        return 0;
    }
    int index = 0;
    for (int i = 0; i < nums.size(); ++i) {
        //如果找到OK的元素，填上坑，言外之意是对于需要删除的元素的坑，无所谓谁去填
        if (nums[i] != val) { 
            nums[index++] = nums[i];
        }
    }
    return index;
}