[leetcode] Two Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/Otori/article/details/40862119

本文介绍了一种经典的算法问题——寻找数组中两个数相加等于特定目标值的索引。给出了两种解决方案：一种是超时的暴力求解法，另一种是通过使用哈希表进行优化的方法。

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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

（超时）暴力求解方法：

	public int[] twosum(int[]numbers, int target){
		int[] result=new int[2];
		int len=numbers.length;
		int sum;
		
		for(int i=0;i<len-1;i++){
			for(int j=i+1;j<len;j++){
				sum=numbers[i]+numbers[j];
				if(sum==target){
					result[0]=i+1;
					result[1]=j+1;
				}
			}
		}
		return result;
		
	}

利用Hashmap，首先遍历数组，将numbers[index]=value, 存入hashmap<value,index>；然后遍历数组，在hashmap中寻找对应的值。

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
		int[] result=new int[2];
		int len=numbers.length;
		Map<Integer, Integer> map = new HashMap<Integer, Integer>();
		
		for(int i=0;i<len;i++){
			map.put(numbers[i], i);
		}
		
		for(int one=0;one<len;one++){
			Integer two=map.get(target-numbers[one]);		
			if(two!=null && one<two){
					result[0]=one+1;
					result[1]=two+1;
				}
			}
		return result;
    }
}

In addtion:

Map<Integer,Integer> map=new HashMap<Integer,Integer>(); //新建map

map.put(numbers[i],i) // put(key,value) 存入键值

map.get(numbers[i]); //get(key)取值