1.tow sum

题目

1. Two Sum

Easy

Given 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, and you may not use the same element twice.

Example:

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

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

c++版本

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ivec;
        for(int i = 0; i < nums.size() - 1; i++){
            for(int j = i + 1; j < nums.size(); j++)
                if(nums[i] + nums[j] == target){
                    ivec.push_back(i);
                    ivec.push_back(j);
                    break;
                }
        }
        return ivec;
    }
};

java版本

class Solution {
    public int[] twoSum(int[] nums, int target) {
            int[] ret = new int[2];
             for(int i = 0; i < nums.length; i++){
             for(int j = i + 1; j < nums.length; j++){
                 if(nums[i] + nums[j] == target){
                     ret[0] = i;
                     ret[1] = j;
                     break;
                 }
             }
         }
        return ret;
    }
}

时间复杂度为O(n^2)

利用hashmap

错误做法

改进

正确做法

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] ret = new int[2];
        HashMap<Integer,Integer> map = new HashMap<>();
     
        for(int i = 0; i <  nums.length; i++){
            int diff =  target - nums[i];
            if(map.containsKey(diff)){
                ret[0] = i;
                ret[1] = map.get(diff);
                break;
            }
            map.put(nums[i],i);//以nums[i]为键
        }
        return ret;
    }
}

时间复杂度为O(n)