1. Two Sum【数组|哈希表】

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.

 

版本1：O(n^2)  【暴力 原始版本】O（1）

class Solution(object):
    def twoSum(self, nums, target):
        length = len(nums)
        for i in range(length):
            for j in range(i+1,length)://游标后移
                if nums[i] + nums[j] == target:
                    return [i,j]

　　

版本2：O(n^2)  【暴力 枚举函数增强】O（1）

class Solution(object):
    def twoSum(self, nums, target):
        for index , item in enumerate(nums):
            for past_index , past_item in enumerate(nums[index+1:],index+1):
                if item + past_item == target:
                    return [index, past_index]

　　

版本3：O(n)    【双程hash table】O（n）

class Solution(object):
    def twoSum(self, nums, target):
        dd = dict()
        for index , value in enumerate(nums):
            dd[value] = index
        for index , value in enumerate(nums):
            tt = target - value
            if dd.has_key(tt) and dd[tt] != index:
                return [index , dd[tt]]

　　

版本4：O(n)    【单程hash table】O（n）

Python

class Solution(object):
    def twoSum(self, nums, target):
        dd = dict()
        for index , value in enumerate(nums):
            tt = target - value
            if dd.has_key(tt) and dd[tt] != index:
                return [dd[tt],index]
            else:
                dd[value] = index

　　

Java

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]) ){
        		map.put(nums[i], i );
        	}else{
        		int[] rs = { map.get(target-nums[i]) , i };
        		return rs;
        	}
        }
        return nums;
    }

 

1.enumerate内置函数的使用（枚举函数）---用于既要遍历索引又要遍历元素；可以接收第二个参数用于指定索引起始值。