LeetCode 1. Two Sum

原题

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].

解题思路

暴力：双循环算法遍历所有情况，时间复杂度为O（n^2），提交后落后99%的人。。。

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        ret = []
        length = len(nums)
        for i in xrange(length):
            for j in xrange(i):
                if nums[i] + nums[j] == target:
                    ret.append(j)
                    ret.append(i)
                    return ret

　　

优化：空间换时间，将遍历过的数都用字典存下来，一旦碰到跟前面的数相加等于target时则返回，这下时间复杂度只有O（n）啦，提交后领先99%的人。。爽啊~~~

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        ret = [0]*2
        temp = {}
        length = len(nums)
        for i in xrange(length):
            if (temp.has_key(target - nums[i])):
                ret[1] = i
                ret[0] = temp[target - nums[i]]
                return ret
            temp[nums[i]] = i