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Commit 16281aa

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增加题目59
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# 删除排序数组中的重复项 II
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#### *Remove Duplicates from Sorted Array II*
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给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素最多出现两次,返回移除后数组的新长度。
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不要使用额外的数组空间,你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。
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说明:
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为什么返回数值是整数,但输出的答案是数组呢?
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请注意,输入数组是以"引用"方式传递的,这意味着在函数里修改输入数组对于调用者是可见的。
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你可以想象内部操作如下:
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// nums 是以"引用"方式传递的。也就是说,不对实参做任何拷贝
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int len = removeDuplicates(nums);
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// 在函数里修改输入数组对于调用者是可见的。
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// 根据你的函数返回的长度, 它会打印出数组中该长度范围内的所有元素。
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for (int i = 0; i < len; i++) {
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print(nums[i]);
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}
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英文题目:
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Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
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Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
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**example 1**
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```
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Given nums = [1,1,1,2,2,3],
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Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
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It doesn't matter what you leave beyond the returned length.
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```
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**example 2**
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```
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Given nums = [0,0,1,1,1,1,2,3,3],
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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.
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It doesn't matter what values are set beyond the returned length.
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```
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---
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### 思路
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1. 遍历数组,index从零开始,使用count记录每个元素出现次数,大于2次则忽略,小于2次则赋值到index的位置,同时index++
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2. 遍历完毕返回index+1则为数组长度
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3. 同时附上评论区大神思路
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### 代码
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```
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#include <vector>
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#include <iostream>
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using namespace std;
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class Solution {
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public:
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int removeDuplicates(vector<int>& nums) {
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int count = 0;
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int index = 0;
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for (int i = 0; i < nums.size(); i++)
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{
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if (index < 1 || nums[i] == nums[index - 1] && count < 2)
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{
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nums[index++] = nums[i];
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++count;
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}
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else if (nums[i] != nums[index - 1])
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{
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nums[index++] = nums[i];
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count = 1;
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}
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}
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return index;
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}
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};
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//大神思路
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int removeDuplicates(vector<int>& nums) {
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int i = 0;
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for (int n : nums)
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if (i < 2 || n > nums[i-2])
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nums[i++] = n;
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return i;
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}
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```
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本题以及其它leetcode题目代码github地址: [github地址](https:github.com/SherlockUnknowEn/leetcode)
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/************************************************************************/
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/*
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Remove Duplicates from Sorted Array II
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Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
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Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
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Example 1:
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Given nums = [1,1,1,2,2,3],
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Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
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It doesn't matter what you leave beyond the returned length.
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Example 2:
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Given nums = [0,0,1,1,1,1,2,3,3],
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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.
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It doesn't matter what values are set beyond the returned length.
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Clarification:
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Confused why the returned value is an integer but your answer is an array?
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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.
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Internally you can think of this:
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// nums is passed in by reference. (i.e., without making a copy)
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int len = removeDuplicates(nums);
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// any modification to nums in your function would be known by the caller.
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// using the length returned by your function, it prints the first len elements.
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for (int i = 0; i < len; i++) {
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print(nums[i]);
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}
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*/
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/************************************************************************/
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#include <vector>
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#include <algorithm>
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#include <iostream>
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using namespace std;
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class Solution {
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public:
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int removeDuplicates(vector<int>& nums) {
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int count = 0;
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int index = 0;
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for (int i = 0; i < nums.size(); i++)
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{
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if (index < 1 || nums[i] == nums[index - 1] && count < 2)
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{
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nums[index++] = nums[i];
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++count;
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}
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else if (nums[i] != nums[index - 1])
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{
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nums[index++] = nums[i];
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count = 1;
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}
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}
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return index;
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}
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};
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int main()
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{
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vector<int> nums = { 1,1,1,2,2,3 };
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int len = Solution().removeDuplicates(nums);
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for (int i = 0; i < len; ++i)
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{
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cout << nums[i] << " ";
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}
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return 0;
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}

‎README.md‎

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##### 55. [Search a 2D Matrix](https://github.com/SherlockUnknowEn/leetcode/tree/master/50-59/55.%20Search%20a%202D%20Matrix(Medium)) 搜索二维矩阵
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##### 56. [Sort Colors](https://github.com/SherlockUnknowEn/leetcode/tree/master/50-59/56.%20Sort%20Colors(Medium)) 颜色排序
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##### 57. [Subsets](https://github.com/SherlockUnknowEn/leetcode/tree/master/50-59/57.%20Subsets(Medium)) 子集
64-
##### 58. [Word Search](https://github.com/SherlockUnknowEn/leetcode/tree/master/50-59/58.%20Word%20Search(Medium)) 单词搜索
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##### 58. [Word Search](https://github.com/SherlockUnknowEn/leetcode/tree/master/50-59/58.%20Word%20Search(Medium)) 单词搜索
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##### 59. [Remove Duplicates from Sorted Array II](https://github.com/SherlockUnknowEn/leetcode/tree/master/50-59/59.%20Remove%20Duplicates%20from%20Sorted%20Array%20II(Medium)) 删除排序数组中的重复项 II

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