27. Remove Element

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.
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?

----分析：-----------------------------------------------------------------

1. 题目要求移除整型数组中与某值相等的所有元素，并返回 int length;

2. 在不分配额外的内存(Do not allocate extra space for another array)，如何移除数组中的与某值相等的元素？

----思路：-----------------------------------------------------------------

可通过设置一个初始下标int index = 0；把不等于val的值赋值给nums[index]，然后index++；

如果nums[i]==val， index不自增1，++i。具体思路见下面图片：

代码如下：

Code1： 时间复杂度：O(n)，空间复杂度 O(1)

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

Code2：时间复杂度：O(n)，空间复杂度 O(1)

借助 STL中的remove， 直接返回：

return distance(nums.begin(), remove(nums.begin(), nums.end(), val )  );

PS：具体的STL的remove方法见这篇blog 

STL移除型算法remove

。