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  • leetcode-1-Two Sum

    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");
        }
    }
    
    
    
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  • 原文地址:https://www.cnblogs.com/vitasoy/p/11875598.html
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