1-找出数组中重复的数字

//把每个数放到对应的位置=>i=nums[i]
class Solution {
public:
    int duplicateInArray(vector<int>& nums) {
        int n = nums.size();
        for (auto x : nums)
            if (x < 0 || x >= n)
                return -1;
        for (int i = 0; i < n; i ++ ) {
            while (nums[nums[i]] != nums[i]) swap(nums[i], nums[nums[i]]);
            if (nums[i] != i) return nums[i];
        }
        return -1;
    }
};

 1  1class Solution {
 2  2 public:
 3  3     int findRepeatNumber(vector<int>& nums) {
 4  4         int n=nums.size();
 5  5         bool mark[n];
 6  6         memset (mark,0,sizeof mark);//*
 7  7         for(auto x:nums) if(x<0||x>n-1) return -1;
 8  8         for(int i=0;i<n;i++)
 9  9         {
10 10             if(mark[nums[i]]) return nums[i];
11 11             else mark[nums[i]]=1;
12 12         }
13 13         return -1;
14 14     }
15 15     }
16 16 };

 1 //计数时判定，若当前数字之前出现过则返回
 2 class Solution {
 3 public:
 4     int findRepeatNumber(vector<int>& nums) {
 5         int n=nums.size();
 6         vector<int>cnt (n,0);
 7         for(auto x:nums) if(x<0||x>n-1) return -1;
 8         for(int i=0;i<n;i++)
 9         {
10             if(cnt[nums[i]]++) return nums[i];
11         }
12         return -1;
13     }
14 };