1. Two Sum

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, and you may not use the same element twice.

Example:

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

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


法1 ：暴力

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

法2：利用dict 映射，
遍历一遍list,将target减去当前元素的差作为key，当前元素的下标为val
如果当前元素是字典的key ,则返回当前元素的下标，与字典中当前元素对应的val

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        n_dict = {}
        for i in range(len(nums)):
            if nums[i] in n_dict.keys():
                return [n_dict[nums[i]],i]
            else:
                n_dict[target-nums[i]] = i;
        

20180325
用 key in dict: 判断效率更高！！！
时间复杂度 o(n)
空间复杂度 o(n)

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        n_dict = {}
        for i,item in enumerate(nums):
            val = target - item
            if item in n_dict:
                return [n_dict[item],i]
            else:
                n_dict[val] = i