leetCode: 1.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:

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Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution

First

45ms

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class Solution {
    public int[] twoSum(int[] nums, int target) {
        int len = nums.length;
        int i=0,j=0;
        for(;i<len;i++){
            for(j=i+1; j<len;j++){
                if(nums[i]+nums[j] == target){
                    return new int[]{i,j};
                }
            }
        }
        return null;
    }
}

Better

7ms

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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++){
            if(map.containsKey(target - nums[i])){
                return new int[]{map.get(target - nums[i]),i};
            }
            map.put(nums[i],i);
        }
        return new int[]{0,0};
    }
}

Faster

说是6ms.. 记录下

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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++){
            if(map.get(target - nums[i])!=null){
                return new int[]{map.get(target - nums[i]),i};
            }
            map.put(nums[i],i);
        }
        return new int[]{0,0};
    }
}

Note

擅于利用map搜索