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Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example,
Given input array A = [1,1,2],
Your function should return length = 2, and A is now [1,2].
思路：
题目不难。直接贴答案。
题解：
class Solution { public: int removeDuplicates(int A[], int n) { int res = 1; for(int i=1;i<n;i++) { if(A[i]!=A[i-1]) A[res++] = A[i]; } return res; } };
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array A = [1,1,1,2,2,3],
Your function should return length = 5, and A is now [1,1,2,2,3].
思路：
和第一题差不多，多用一个变量表示重复的次数。
题解：
class Solution { public: int removeDuplicates(int A[], int n) { if(n==0) return 0; int repeated = 0; int res = 1; for(int i=1;i<n;i++) { if(A[i]!=A[i-1]) { A[res++] = A[i]; repeated = 0; } else { if(repeated<1) A[res++] = A[i]; repeated++; } } return res; } };
[leetcode] Remove Duplicates from Sorted Array I II
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原文地址：http://www.cnblogs.com/jiasaidongqi/p/4159157.html
