LeetCode--283--移动0

问题描述：

给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。

示例:

输入: [0,1,0,3,12]
输出: [1,3,12,0,0]

说明:

1. 必须在原数组上操作，不能拷贝额外的数组。
2. 尽量减少操作次数。

方法1：检测到为0纪录0的个数，不为0时进行赋值运算nums[i - k] = nums[i]

　　　　最后将检测到的0补到nums的最后

 class Solution(object):
     def moveZeroes(self, nums):
         """
         :type nums: List[int]
         :rtype: void Do not return anything, modify nums in-place instead.
         """
         if len(nums) == 2:
             if nums[0] == 0:
                 nums[0],nums[1] = nums[1],nums[0]
                 return
             else:
                 return
         k = 0
         for i in range(len(nums)):
             if nums[i] == 0:
                 k += 1
             elif nums[i] != 0 :
                 nums[i - k] = nums[i]
         for j in range(len(nums) - k,len(nums)):
             nums[j] = 0

方法2：每次将元素等于0的位置变成[]。

 class Solution(object):
     def moveZeroes(self, nums):
         """
         :type nums: List[int]
         :rtype: void Do not return anything, modify nums in-place instead.
         """
         i = 0
         length = len(nums)
         while i < len(nums):
             while nums != [] and nums[i] == 0:
                 nums[i:i+1] = []
                 if i == len(nums):
                     break
             i += 1
         nums += (length - len(nums))*[0]

将方法1改进:

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

2018-09-23 09:07:49