283. Three solutions for moving zero double pointers

Title description

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

Example:

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

Description :

1. You must operate on the original array, and cannot copy additional arrays.
2. Minimize the number of operations.

 Solution 1: Double pointer, the first 0 is exchanged with the first non-zero after it

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int len = nums.size();
        int first0 = -1;    //无效的初始index

        for (int i = 0; i < len; ++i)
            if (0 == nums[i]) {
                if (first0 < 0)              //寻找第一个0的index
                    first0 = i;
            } else if (first0 >= 0) {        //第一个0已找到，且找到第一个0后面第一个非0元素
                swap(nums[first0], nums[i]); //把第一个0换成非0元素
                i = first0;                  //从刚才第一个0后面重新找新的第一个0
                first0 = -1;
           }
    }
};

Solution 2: Double pointer, j records the index where the non-zero element should be, i finds the non-zero element; there is a repeated operation of self-exchange

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int len = nums.size();

        int j = 0;                        //遇非0元素自加
        for (int i = 0; i < len; ++i)
            if (nums[i])
                swap(nums[j++], nums[i]); //把非0元素全部换到前面
    }
};

Solution 3: Optimization of solution 2 to avoid exchanging values ​​with yourself

//如果元素不为0，i,j跟着一起动。如果元素为零，j记录为零的下标
class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int len = nums.size();

        int j = 0;
        for (int i = 0; i < len; ++i)
            if (nums[i]) {
                if (i > j) {
                    nums[j] = nums[i];
                    nums[i] = 0;
                }
                ++j;
            }
    }
};