标题叙述性说明:
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]
.
解题思路:遍历数组。用count记录当前遇到反复元素的对数,将每一个元素向前移动count位。
代码:
int Solution::removeDuplicates(int A[], int n) { int count = 0; for(int i = 0;i < n;i++) { if(i != 0) { if(A[i] == A[i-1]) count++; A[i-count] = A[i]; } } return n-count; }
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