题目

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.

分析

直接使用STL中的unique和erase函数即可。

实现一

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        nums.erase(unique(nums.begin(), nums.end()), nums.end());
        return nums.size();
    }
};

思考

感觉直接用STL有投机取巧的嫌疑，所以再写一个手工的版本。

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.empty()) return 0;
        int len=1;
        for(int i=1; i<nums.size(); i++){
            if(nums[len-1]!=nums[i]){
                swap(nums[len], nums[i]);
                len++;
            }
        }
        return len;
    }
};