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]
.
这题和remove element有点类似,只是这里需要把数组中所有重复的数都只留下一个。我先借鉴了那道题的思路,代码如下:
class Solution { public: int removeDuplicates(int A[], int n) { int k = 0; for(int i = 1; i < n; i++) { if(A[k] != A[i]) { k++; } else { for(int j = i; j < n-1; j++) A[j] = A[j+1]; n--; i--; } } return n; } };
class Solution { public: int removeDuplicates(int A[], int n) { if(n == 0) return 0; int k = 1; for(int i = 1; i < n; i++) { if(A[i] == A[i-1]) continue; else { A[k] = A[i]; k++; } } return k; } };改进后的代码如上,时间复杂度减少了一维,遍历一次数组即可。
Leetcode13: Remove Duplicates from Sorted Array
原文:http://blog.csdn.net/u013089961/article/details/45154997