LeetCode 283. Move Zeroes [two pointers]

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:

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

 

题目要求保持非零数字相对位置不变的前提下，把0移到末尾

最开始的思路是遇见0之后，把所有的数字都向前移动一位，通过了大部分测试用例，但是无法处理连零的问题，代码如下：

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int size = nums.size();
        for(int i=0;i<size;i++)
        {
            if(nums[i]==0)
            {
                for(int j=i+1;j<size;j++)
                   nums[j-1]=nums[j];
                nums[size-1]=0;
            }
        }
        
    }
};

换个思路，遍历两遍向量，记录下来0的数量，将非零数字存入新的向量，最后再将0补进新的向量，时间空间复杂度均是O(n)，代码如下：

void moveZeroes(vector<int>& nums) {
    int n = nums.size();

    // Count the zeroes
    int numZeroes = 0;
    for (int i = 0; i < n; i++) {
        numZeroes += (nums[i] == 0);
    }

    // Make all the non-zero elements retain their original order.
    vector<int> ans;
    for (int i = 0; i < n; i++) {
        if (nums[i] != 0) {
            ans.push_back(nums[i]);
        }
    }

    // Move all zeroes to the end
    while (numZeroes--) {
        ans.push_back(0);
    }

    // Combine the result
    for (int i = 0; i < n; i++) {
        nums[i] = ans[i];
    }
}

进一步优化，用两个指针，第一个指针记录最后一个非零元素应该在的位置，第二个指针负责遍历，遇见非零元素就把发赋值给第一个元素所在位置，然后第一个指针前进一个，第二次遍历则是把第一个指针到向量的末尾全部赋值为0。时间复杂度O（n），空间复杂度O(1)；

void moveZeroes(vector<int>& nums) {
    int lastNonZeroFoundAt = 0;
    // If the current element is not 0, then we need to
    // append it just in front of last non 0 element we found. 
    for (int i = 0; i < nums.size(); i++) {
        if (nums[i] != 0) {
            nums[lastNonZeroFoundAt++] = nums[i];
        }
    }
    // After we have finished processing new elements,
    // all the non-zero elements are already at beginning of array.
    // We just need to fill remaining array with 0's.
    for (int i = lastNonZeroFoundAt; i < nums.size(); i++) {
        nums[i] = 0;
    }
}

再优化操作方面，只遍历一遍，两个指针，思路同上，这次赋值变成交换位置，交换的次数为非零元素的个数

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