Leetcode OJ 1Two Sum

https://leetcode.com/problems/two-sum/

Water problem, but since retired very few do the problem, really backward too strong, not consider the whole

Test instructions: Give an array, also a target, and ask which two numbers to add up to get target

Answer: Bucket row Orhash

1, note that the bucket sort, and the depth of the bucket is not necessarily 1, so hash[i] represents the number of I rather than is not present

2, because the subscript is involved, so be careful that the number of arrays can be fractions, my approach is to find min (X[i]), if min (X[i]) <0, then target+=-2*min (X[i]), x[i]+= ( -1) *min (x[i ]),

Very not accustomed to IS, leetcode OJ incredibly does not have a given number of range, this very headache, the array open how big ...

Class Solution {public:vector<int> Twosum (vector<int>& nums, int target) {int tmp,cnt=0;        int num[100001],id[100001];        int t = 0;        memset (num,0,sizeof (num));        Memset (id,0,sizeof (id));            for (int i=0;i<nums.size (); i++) {tmp = nums[i];        t = min (t,tmp);        } cnt=0;            if (t<0) {t=-t;                for (int i=0;i<nums.size (); i++) {nums[i] + = t;                int tmp= Nums[i];                NUM[TMP] + +;                cnt++;                id[tmp]=cnt;            cout << nums[i] << Endl;        } target + = 2*t;                }else{for (int i=0;i<nums.size (); i++) {int tmp= nums[i];                NUM[TMP] + +;                cnt++;                id[tmp]=cnt;            cout << nums[i] << Endl; }}//cout << "ttt=" << target << "<< t << endl        int flag=0;        vector<int>ans;                    for (int i=0;i<cnt;i++) {if (nums[i]<=target) {if (Num[target-nums[i]]) {                    if (target-nums[i] = = Nums[i] && num[nums[i]]<2) continue;                    flag = 1;                    Ans.push_back (i+1);                    Ans.push_back (id[target-nums[i]);                    Ans = i+id[Target-nums[i]];                Break    }}} return ans; }};

Leetcode OJ 1Two Sum