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唐树森 2018-09-01 18:42:01 +08:00 committed by GitHub
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# 思路
由于是已经排序好的数组,所以相同的数肯定是相邻的,遍历数组时跳过与上一个数重复的数即可。
可以用一个count记录在下标count及之前是没有重复的数组 count初始化为0然后从前往后遍历若a[i]==a[count]说明重复不更新count
否则应该让count++。注意单独判断空数组。
# C++
```
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.empty()) return 0;
int count = 0;
for(int i = 1; i < nums.size(); i++){
if(nums[i] != nums[count]) nums[++count] = nums[i];
}
return count+1;
}
};
```