leetcode-1-Two Sum

标签：哈希表   ber   遍历   strong   exce   public   sum   new   etc   

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].

方法一:暴力法

方法二：哈希表

如题目所示，第一个元素是2，则查看哈希表是否有target-2的元素，如果没有则向后遍历，并将第一个元素（2，index)存入哈希表，遍历至7时，查看哈希表(target-7)=2的元素存在，即可以返回map(2)和当前元素所在的下标

代码：

class Solution {
    public int[] twoSum(int[] nums, int target) {
            Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0;i<nums.length;i++){
            int complent = target - nums[i];
            if(map.containsKey(complent)&&map.get(complent)!=i){
                return new int[] {i,map.get(complent)};
            }
            map.put(nums[i], i);
        }
        throw new RuntimeException("No two sum");
    }
}

leetcode-1-Two Sum

标签：哈希表   ber   遍历   strong   exce   public   sum   new   etc   

原文地址：https://www.cnblogs.com/vitasoy/p/11875598.html