Add Two Numbers

Number of 1 Bits

iven an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

For example

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路

最简单的2重循环，自己也是这样想的

代码1 –2重循环

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        for(size_t it = 0;it<nums.size()-1;it++)
        {
            for(size_t it2 = it+1;it2<nums.size();it2++)
            {
                if(nums[it]+nums[it2]==target)
                {
                    res.push_back(it);
                    res.push_back(it2);
                    return res;
                }

            }
        }

    }
};

代码2 –O(n)

用到了map，存nums[x],x
然后就在里面找target-nums[x]
这样找到第二个的时候，第一个肯定是配对好的了

    public int[] twoSum(int[] nums, int target) {

        if(nums==null || nums.length==0)
            return null;
        int[] result = new int[2];
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int x=0;x<nums.length;x++){
            if(map.containsKey(target-nums[x])){
              result[1] = x;
              result[0] = map.get(target-nums[x]);
            }
            map.put(nums[x],x);
        }
        return result;
    } };