leetcode笔记--Move Zeroes



Move Zeroes

题目：难度（Easy）

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.
Tags：Array，Two Pointers

分析：要求：就地更改数组，并且使得所有的操作最小

法1：就统计非0数的个数，缺点：可能遍历2遍数组，时间复杂度O(n)

代码实现：

class Solution(object):    def moveZeroes(self, nums):        """        :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.        """               n = len(nums)        countOfNoZero = 0        for i in range(n):            if nums[i]:                nums[countOfNoZero] = nums[i]                countOfNoZero += 1        for i in range(countOfNoZero, n):            nums[i] = 0

法2：双指针,时间复杂度O(n)

代码实现：

class Solution(object):    #双指针       def moveZeroes(self, nums):        """        :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.        """        n = len(nums)        i, j = 0, 0#i指向第一个0，j指向第一个非0        while j < n and i<n:            while j<n and nums[j]==0:                j += 1            while i<n and nums[i]!=0:                i += 1            if j == n or i == n:                break            if j>i:#只有出现在第一个0的后面的非0数才需要移动，并占据第一个0 的位置                nums[i] = nums[j]                nums[j] = 0            else:#第一个非0数在第一个0的前面，不需要移动，则需要寻找下一个可能需要移动的非0数                j += 1