Leetcode13: Remove Duplicates from Sorted Array
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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]
.
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; }};
但是这里会报超时,因为时间复杂度是n方。其实仔细想想,时间复杂度可以降一维,每次并不需要把删除元素后面的数都往前挪,我们只需要比较相邻两个数(因为是有序的),当不同时把后面的数往前挪即可,用一个计数器来记录索引值。
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。
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