Level:
Easy
题目描述:
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]
思路分析:
给定一个目标值,要返回数组中两个数之和等于该目标值的两数下标,我们可以借助map来实现,将元素作为键,下标作为值存放进map,我们将目标值记为target,当遍历到数组中某个元素num时,我们查看map中是否存有target-num这个键,如果存在那么我们就找到了满足要求的两个数,如果不存在,继续向下遍历,直到找到答案。
代码:
class Solution {
public int[] twoSum(int[] nums, int target) {
int []res=new int[2];
HashMap<Integer,Integer>map=new HashMap<>();
int temp;
for(int i=0;i<nums.length;i++){
temp=target-nums[i];
if(map.get(temp)!=null){
res[0]=map.get(temp);
res[1]=i;
break;
}else{
map.put(nums[i],i);
}
}
return res;
}
}