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 nums = [1,1,2]
,
Your function should return length = 2
, with the first two elements of nums being 1
and 2
respectively.
It doesn't matter what you leave beyond the new length.
思路:用count记录相同的个数。对于nums[i]到它相同的有多少个,那么他实际的位置就应该是nums[i-count];\
代码如下(已通过leetcode)
public class Solution {
public int removeDuplicates(int[] nums) {
int length=nums.length;
int n=0;
for(int i=0;i<length-1;i++) {
if(nums[i]==nums[i+1]) n++;
nums[i-n+1]=nums[i+1];
}
return length-n;
}
}