【Leetcode】1. Two Sum

1. Two Sum

Total Accepted: 254411
Total Submissions: 1031847
Difficulty: Easy

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.

思路：

1、暴力O(n2)

2、hash O(n):
Hash表法，以空间换时间：时间复杂度：O(N)空间复杂度O(N).

更快的查找方法：hash表，给定一个数字，根据hash映射查找另一个数字是否在数组中，只需O（1）的时间，但需要承担O(N)的hash表存储空间。

代码1（暴力）：

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

Runtime: 58 ms

代码2（hash）:

public class Solution {
public int[] twoSum(int[] numbers, int target) {
if (numbers != null) {
HashMap<Integer, Integer> num_map = new HashMap<Integer, Integer>();
for (int i = 0; i < numbers.length; i++) {
if (num_map.containsKey(numbers[i])) {
int index = num_map.get(numbers[i]);
int[] result = { index, i };
return result;
} else {
num_map.put(target - numbers[i], i);
}
}
}
return null;
}
}

Runtime: 6 ms

分析：

遍历传进来的数组int[] numbers，检查target - numbers[i]在不在hash表中，target - number[i]存储为hash表的key。如果在，就返回当前的i，和对应的key。