LeetCode27. Remove Element（模拟）

LeetCode27. Remove Element（数组 / 模拟）

• Easy
• Accepted：360,124
• Submissions：833,153

Description

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]);
}

链接

https://leetcode.com/problems/remove-element/

题意

给定一个数组，删除其中所有值为val的元素，然后返回新数组的长度l，保证新数组中前l个元素的值为非val，顺序可以不一致。和LeetCode283. Move Zeroes（模拟）其实一样。

题解

题目说不允许开辅助空间，那么直接对原数组遍历一遍，遇到非val元素重新赋值即可。

代码

• Runtime: 0 ms, faster than 100.00% of C++ online submissions for Remove Element.

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int len = nums.size();
        int cnt = 0;
        // [0, cnt)中为非val元素，长度为cnt
        for (int i = 0; i < len; ++i)
        {
            if (nums[i] != val)
                nums[cnt++] = nums[i];
        }
        return cnt;
    }
};

• Runtime: 0 ms, faster than 100.00% of C++ online submissions for Remove Element.
• Memory Usage: 8.9 MB, less than 35.69% of C++ online submissions for Remove Element.

class Solution {
private:
    int n, cnt;
public:
    int removeElement(vector<int>& nums, int val) {
        n = nums.size();
        if (n == 0)
            return 0;
        // 统计值为 val 的元素数量
        for (int i = 0; i < n; ++i)
        {
            if (nums[i] == val)
                ++cnt;
        }
        // 把值为 val 的元素换到后面去
        int i = 0, j = n - 1;
        while (i < j)
        {
            if (nums[i] == val && nums[j] == val)
                --j;
            else if (nums[i] == val && nums[j] != val)
            {
                swap(nums[i], nums[j]);
                ++i, --j;
            }
            else
                ++i;
        }
        return n - cnt;
    }   
};