[Leetcode] 27. Remove Element（删除元素）题解

Given an array nums and a value val, 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 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

题目解析：
给定数组和值，删除该值的所有实例并返回新的长度。不要为另一个数组分配额外的空间，您必须使用常量内存来执行此操作。元素的顺序可以改变。无论你离开新的长度什么都不重要。

思路：

1、和26一样的操作，直接利用vector特定的函数erase()来删除给定的元素。

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

[Leetcode] 27. Remove Element（删除元素）题解

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