Remove Duplicates from Sorted Array
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].
主要思路是，有几个不相同的元素，它的下标就是不相同的元素个数减一，因此，当元素相同的时候，下标不变化，但是元素向后移动。
C++代码实现：
#include<iostream> using namespace std; class Solution { public: int removeDuplicates(int A[], int n) { int i; int count=1; //注意数组为空的情况 if(n==0) return 0; for(i=1;i<n;i++) { if(A[i-1]!=A[i]) { //count记录有几个不相等的元素，第一个不相等的元素下标为1，第二个不相等的元素的下标为2，依次类推 count++; A[count-1]=A[i]; } } return count; } }; int main() { int arr[10]={1,2,3,3,3,4,4,5,5,6}; Solution s; cout<<s.removeDuplicates(arr,10)<<endl; for(auto a:arr) cout<<a<<" "; cout<<endl; }
运行结果：
Remove Duplicates from Sorted Array
原文：http://www.cnblogs.com/wuchanming/p/4098061.html