[leetcode] 27. Remove Element

TstsUgeg

于 2016-03-22 10:58:45 发布

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分类专栏： leetcode 文章标签： leetcode Two Pointers

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本文介绍了一种高效的算法，用于删除数组中指定值的所有实例，并返回新的有效长度。该方法使用了两个指针技巧，在原地修改数组，且仅使用常数额外空间。

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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?

这道题是删除数组中等于目标值的元素，题目难度为Easy。

题目比较简单，遍历数组将不等于目标值的元素从数组头部开始逐个写入数组，最终下标位置即是新数组长度，具体代码：

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