LeetCode--Two Sum

问题描述

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.

主要采用的思想

采用hash表的思想，用一个hash函数把每一个值映射到表上
其中hash函数为 h(n) = target-n+SHIFT
其中n为数组中的一个数，target为目标和，SHIFT为一个偏移量（之所以要有偏移是因为数组没有负数的下标，而target-n可能为负数）

代码

public class Solution {    public int[] twoSum(int[] nums, int target) {        int[] result = new int[2];        //get the max and min number of array nums[]        int length = nums.length;        int max=nums[0];        int min=nums[0];        for(int j=1; j<length; j++){            if(nums[j]>max)                max = nums[j];            else if(nums[j]<min)                min = nums[j];        }        //build the hash table        int htableLen;        final int SHIFT;        if(min < 0){            htableLen = max-min+1;            SHIFT = -min;        }else{            htableLen = max+1;            SHIFT = 0;        }        int[] htable = new int[htableLen];        for(int j=0; j<htableLen; j++)//initialize the hashtable            htable[j] = -1;        int hashV;        int num;        for(int i=0; i<length; i++){            num = nums[i];            hashV = target - num + SHIFT;            if(htable[num + SHIFT] != -1){//means that the peer number is here                result[0]=htable[num + SHIFT];                result[1]=i;                return result;            }else if(hashV>=0&&hashV<htableLen){//be sure that the hash value is in the bound of hash table                htable[hashV] = i;            }        }        return result;    }}

运行结果

用大量空间换取时间，速度挺感人的。。。