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.

题解与理解

题意大致是：给定一个数组，和一个目标数字target；找出数组中和为目标数字target的两个数

解题用了一个HashMap；key为该数字target-nums[i]，value为nums[i];

对数组进行遍历，如果map中不存在与其相等的key，则将该数字存入map；

当遍历的数字发现map中存在与其对应的key时，则说明当前遍历数字与前面遍历过的某数字相加为target；
此时获取map中的该数字，存储到result中，并return；

解题源码

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

1.Two Sum

题解与理解

解题源码

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