[leetcode] 80. Remove Duplicates from Sorted Array II

[leetcode] 80. Remove Duplicates from Sorted Array II

Description

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,1,2,2,3],Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// 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]);}

分析

题目的意思是：给定一个数组，如果数字出现频率超过两次，就保留两次，把剩下的移除，然后返回移除数字之后的数组的长度。

代码

class Solution {public: int removeDuplicates(vector& nums) { int m=nums.size(); if(m<=2){ return m; } int pre=0; int cur=1; int count=1; while(cur

参考文献

​​[编程题]remove-duplicates-from-sorted-array-ii​​​​[LeetCode] Remove Duplicates from Sorted Array II 有序数组中去除重复项之二​​

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