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].
这道题很简单，就是维护一个新数组的尾指针，tail，把去重后的数放到A[tail]就行，最终返回tail。
1 class Solution { 2 public: 3 int removeDuplicates(int A[], int n) { 4 if (n <= 1) return n; 5 int tail = 0, v; 6 7 for (int i = 0; i < n;) { 8 v = A[i]; 9 for (i++; i < n && A[i] == v; ++i); 10 A[tail++] = v; 11 } 12 return tail; 13 } 14 };
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].
同上，区别在于，去重的时候，允许有两个重复的数；多的话还是略掉。
1 class Solution { 2 public: 3 int removeDuplicates(int A[], int n) { 4 if (n <= 1) return n; 5 int tail = 0, v, j = 0; 6 7 for (int i = 0; i < n;) { 8 v = A[i]; 9 for (j = 0; j < 2 && i + j < n && A[i + j] == v; ++j) { 10 A[tail++] = v; 11 } 12 for (i += j; i < n && A[i] == v; ++i); 13 } 14 return tail; 15 } 16 };
哈哈，Leetcode第一遍至此结束了！！！！cheers！
Leetcode | Remove Duplicates from Sorted Array I && II,布布扣,bubuko.com
Leetcode | Remove Duplicates from Sorted Array I && II
原文：http://www.cnblogs.com/linyx/p/3735563.html