python将数组的$去掉 python去掉数组中的重复数字

本文实例讲述了Python实现删除排序数组中重复项的两种方法。分享给大家供大家参考，具体如下：

对于给定的有序数组nums，移除数组中存在的重复数字，确保每个数字只出现一次并返回新数组的长度

注意：不能为新数组申请额外的空间，只允许申请O（1）的额外空间修改输入数组

Example 1:
Given nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn't matter what you leave beyond
Example 2:
Given nums = [0,0,1,1,1,2,2,3,3,4],Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
It doesn't matter what values are set beyond the returned length.

说明：为什么返回列表长度而不用返回列表？因为列表传入函数是以引用的方式传递的，函数中对列表进行的修改会被保留。

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

1. 简单判断列表中元素是否相等，相等就删除多余元素

def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
if len(nums)==1: #单独判断列表长度为1的情况，因为之后的for循环从下标1开始
return 1
temp_num = nums[0]
count =0 #for循环中动态删除列表元素，列表缩短，为了防止下标溢出需要用count标记删除元素个数
for index, num in enumerate(nums[1:]):
if temp_num == num: #元素相等就删除
del nums[index-count]
count += 1
else:
temp_num = num
return len(nums)
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
forth = 0
back = 1
while back <= len(nums)-1:
if nums[forth] == nums[back]:
nums.pop(back)
else:
forth += 1
back += 1
return len(nums)

2. 修改数组，保证数组的前几个数字互不相同，且这几个数字的长度同返回长度相等

def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
length = 0 #不存在重复数字的数组长度
for index in range(1,len(nums)): #遍历数组
if nums[index] != nums[length]:
length += 1
nums[length] = nums[index]
return length+1

算法题来自：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/description/

PS：本站还有两款比较简单实用的在线文本去重复工具，推荐给大家使用：

希望本文所述对大家Python程序设计有所帮助。