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  • leetCode-Two Sum

    Description:
    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].

    My Solution:

    class Solution {
        public int[] twoSum(int[] nums, int target) {
            for(int i = 0;i < nums.length;i++){
                for(int j = i + 1;j<nums.length;j++){
                    if((nums[i] + nums[j]) == target){
                        return new int[]{i,j};
                    }
                }            
            }
            throw new IllegalArgumentException("no two sum solution!");
        }
    }

    Better Solution:

    class Solution {
        //用hashMap存储{target - nums[i]:i},如果包含target-nums[i]这个key,说明找到两元素之和为target
        public int[] twoSum(int[] nums, int target) {
            Map<Integer, Integer> map = new HashMap<>();
            for(int i = 0; i < nums.length; i++) {
                if(map.containsKey(nums[i])) {
                    return new int[]{map.get(nums[i]), i};
                } 
                map.put(target - nums[i], i);
            }
            return new int[2];
        }
    }
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  • 原文地址:https://www.cnblogs.com/kevincong/p/7900352.html
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