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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]
.
这题与Remove Duplicates from Sorted List类似,只不过换成了数组,要把排好序的数组的元素变为不重复,且要求原地操作。
class Solution { public: int removeDuplicates(int A[], int n) { int new_arr_cur, arr_cur; new_arr_cur = arr_cur = 0; while (arr_cur < n) { A[new_arr_cur++] = A[arr_cur]; // assign //skip duplicates while (arr_cur < n && A[arr_cur] == A[new_arr_cur - 1]) { arr_cur++; } } return new_arr_cur; } };
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]
.
这里是最多允许两次重复,要在上面的代码进行修改。
class Solution { public: int removeDuplicates(int A[], int n) { int new_arr_cur, arr_cur, count; new_arr_cur = arr_cur = 0; const int DUPLICATE_NUM = 2; while (arr_cur < n) { A[new_arr_cur++] = A[arr_cur++]; count = 1; //add allowed duplicates while (count < DUPLICATE_NUM && arr_cur < n && A[arr_cur] == A[new_arr_cur - 1]) { A[new_arr_cur++] = A[arr_cur++]; count++; } //skip leftmost duplicate while (arr_cur < n && A[arr_cur] == A[new_arr_cur - 1]) { arr_cur++; } } return new_arr_cur; } };
Remove Duplicates from Sorted Array
标签:style blog http color os io for ar art
原文地址:http://blog.csdn.net/freeliao/article/details/38778257