LeetCode283. Move Zeroes（思维）

• Easy
• Accepted：407,019
• Submissions：763,123

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Example

Input: [0,1,0,3,12]
Output: [1,3,12,0,0]

Note

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

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链接

https://leetcode.com/problems/move-zeroes/

题意

给定一个数组，将其中所有为0的元素移到数组的末尾并保持非0元素的相对位置不变。

题解

• 直接暴力，统计一下0的个数再开一个辅助数组就行，时间复杂度$O(n)$，空间复杂度$O(n)$
• 对原数组进行遍历，先将所有的非0元素移到数组前面，最后再加上0元素，时间复杂度$O(n)$，空间复杂度$O(1)​$

代码

• Runtime: 8 ms, faster than 100.00% of C++ online submissions for Move Zeroes.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        vector <int> ans;
        int len = nums.size();
        int cnt = 0;
        for (int i = 0; i < len; ++i)
        {
            if (nums[i] != 0)
                ans.push_back(nums[i]);
            else
                cnt++;
        }
        while (cnt--)
            ans.push_back(0);
        for (int i = 0;i < len; ++i)
            nums[i] = ans[i];
    }
};

• Runtime: 8 ms, faster than 100.00% of C++ online submissions for Move Zeroes.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int len = nums.size();
        int cnt = 0;
        for (int i = 0; i < len; ++i)
        {
            if (nums[i] != 0)
                nums[cnt++] = nums[i];
        }
        for (int i = cnt; i < len; ++i)
            nums[cnt++] = 0;
    }
};

• 对原数组进行遍历，找到非0元素时，交换位置，最后无需再加上0元素

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int len = nums.size();
        int cnt = 0; // [0, cnt)中均为非0元素
        for (int i = 0; i < len; ++i)
        {
            if (nums[i] != 0)
                swap(nums[cnt++], nums[i]);
        }
    }
};

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The end.
2019年2月1日 星期五