题目

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

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

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

题目给出了一个数组，找出其中的两个数之和等于给定的数，并返回两个数的下标（从1开始）。

利用HashMap从左往右扫描一遍，然后将数及坐标，存到map中，扫描过程中判断目标值和当前扫描值之差是否在map中，然后返回相应的下标即可。时间复杂度O(n)。代码如下：

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        Map<Integer,Integer> map=new HashMap<Integer,Integer>();
        int len=numbers.length;
        int[] results=new int[2];
        for(int i=0;i<len;i++){
            if(map.containsKey(target-numbers[i])){
                results[0]=map.get(target-numbers[i])+1;  //注意因为返回的结果下标是从1开始的，所以要加1
                results[1]=i+1;  
            }else{
                map.put(numbers[i],i);
            }
        }
        return results;
    }
}

//法2：暴力法，超时
public static int[] twoSum(int[] numbers,int target){
	int[] ret=new int[2];
	int len=numbers.length;
	for(int i=0;i<len-1;i++){
		for(int j=i+1;j<len;j++){
			if(numbers[i]+numbers[j]==target){
				ret[0]=i+1;
				ret[1]=j+1;
			}
		}
	}
	return ret;
}

---EOF---

【LeetCode】Two Sum