leetcode-1. Two Sum

前言：刷了几天简单的题找感觉，现在决定按照序号来做题。

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.

Example:

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

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

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

首先想到的一个算法是O(n^2)的时间复杂度。Your runtime beats 19.75% of javasubmissions

public class Solution {
    public int[] twoSum(int[] nums, int target) {
	int i=0,j=1;
	boolean flag = true;
        for(; i < nums.length - 1; i ++){
		
		for( j = i + 1; j < nums.length; j ++){
			if(target == (nums[i] + nums[j])){
				flag = false;
				break;
			}
		}
		if(!flag) break;
	}
	return new int[]{i,j};
        
    }
}

接下来思考的是，第二个循环是不是可以省略，也就是是否存在可能我知道差的那个数在哪，那么自然想到可以先遍历一遍将所有数据存到一个HashMap里面就可以了。

时间复杂度O(n);

Your runtime beats 41.62% of javasubmissions.

public class Solution {
    public int[] twoSum(int[] nums, int target) {

	int i=0,j=0;
	HashMap<Integer,Integer> nums_map = new HashMap<Integer,Integer>();	
	for( ; i < nums.length ; i ++){
		if(nums_map.containsKey(nums[i])){nums_map.put(nums[i],nums_map.get(nums[i]) + i);}
		else{nums_map.put(nums[i],i);}
	}

        for(i =0 ; i < nums.length; i ++){
		int remainder = target - nums[i];
		if(nums_map.containsKey(remainder)){
			j = nums_map.get(remainder);
			if(remainder == nums[i]) j -= i;
			if(j != 0)
				break;
		}
		
	}
	return new int[]{i,j};
        
    }
}

接下来继续思考，是否只要循环一遍就可以了，确实还真可以，因为题目知道存在唯一解。因此我们其实遍历一遍就可以了。

Your runtime beats 91.17% of javasubmissions.

public class Solution {
    public int[] twoSum(int[] nums, int target) {

	int i=0,j=0;
	HashMap<Integer,Integer> nums_map = new HashMap<Integer,Integer>();	

        for(; i < nums.length; i ++){
		int remainder = target - nums[i];
		if(nums_map.containsKey(remainder)){
			j = nums_map.get(remainder);
			break;
		}else{
			nums_map.put(nums[i],i);
		}
		
	}
	return new int[]{i,j};
        
    }
}