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。。。