LeetCode-1. Two Sum

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].

暴力，时间复杂度为O(n²)

public int[] twoSum(int[] nums, int target) {//my
        int[] re=new int[2];
        int sign=0;//找到的标记（结束标记）
        for(int i=0;i<nums.length-1;i++){
            if (1==sign){
                break;
            }
            for(int j=i+1;j< nums.length;j++){
                if(nums[i]+nums[j]==target){
                    re[0]=i;
                    re[1]=j;
                    sign=1;
                    break;
                }
            }
        }
        return re;
    }

　　

时间复杂度为O(n)的方法

public int[] twoSum(int[] nums, int target) {//map mytip
        int[] re=new int[2];
        Map<Integer,Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int other = target- nums[i];
            if(map.containsKey(other)){
                re[0] = map.get(other);
                re[1]= i;
            }
            else{
                map.put(nums[i],i);
            }
        }
        return re;
        
    }

 

进阶题

三数之和 LeetCode15 https://www.cnblogs.com/zhacai/p/10579514.html

四数之和 LeetCode18 https://www.cnblogs.com/zhacai/p/10580394.html

剑指offer 和为S的两个数字 https://www.cnblogs.com/zhacai/p/10696698.html