LeetCode--283. Move Zeroes

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本文介绍LeetCode第283题Move Zeroes的两种解决方案：一种是通过删除和添加0的方法；另一种是更优的双指针法，通过交换非零元素来减少操作次数，提高效率。

283. Move Zeroes

题目描述：

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.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:

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

大致的意思就是给一个数组，将数组中所有的0移到数组的末尾，并且保证非零数字的顺序不能变。

代码如下：

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        vector<int>::iterator it=nums.begin();
        for(int i=0;i<nums.size();i++){    //遍历数组一遍
            if(*it==0){
                it=nums.erase(it); //如果为0，将其删除，此时it指向被删除的0的下一个元素
                nums.push_back(0); //向数组的末尾放一个0
            }
            else{      //不是0，比较下一个数值
                it++;
            }
        }
    }
};

提交后，用时22ms，击败了22.22%的C++程序。

LeetCode提供的最佳解法如下：

Approach #3 (Optimal) [Accepted]

The total number of operations of the previous approach is sub-optimal. For example, the array which has all (except last) leading zeroes: [0, 0, 0, ..., 0, 1].How many write operations to the array? For the previous approach, it writes 0's $n - 1$ times, which is not necessary. We could have instead written just once. How? ..... By only fixing the non-0 element,i.e., 1.

The optimal approach is again a subtle extension of above solution. A simple realization is if the current element is non-0, its' correct position can at best be it's current position or a position earlier. If it's the latter one, the current position will be eventually occupied by a non-0 ,or a 0, which lies at a index greater than 'cur' index. We fill the current position by 0 right away,so that unlike the previous solution, we don't need to come back here in next iteration.

In other words, the code will maintain the following invariant:

All elements before the slow pointer (lastNonZeroFoundAt) are non-zeroes.

All elements between the current and slow pointer are zeroes.

Therefore, when we encounter a non-zero element, we need to swap elements pointed by current and slow pointer, then advance both pointers. If it's zero element, we just advance current pointer.

With this invariant in-place, it's easy to see that the algorithm will work.

首先设置一个变量lastNonZeroFoundAt，初值为0，用来记录“最后一个非零数字的下标”，最后是相对来讲的，设置变量cur=0用来遍历数组，cur的值就是当前遍历到的位置。如果nums[cur]为非零元素，则交换nums[lastNonZeroFoundAt]和nums[cur]的值并且lastNonZeroFoundAt++,cur++，如果nums[cur]为零，则只需cur++，写一个数组模拟一下会更容易理解，官方代码如下：

void moveZeroes(vector<int>& nums) {
    for (int lastNonZeroFoundAt = 0, cur = 0; cur < nums.size(); cur++) {
        if (nums[cur] != 0) {
            swap(nums[lastNonZeroFoundAt++], nums[cur]);
        }
    }
}