LeetCode OJ_题解(python):001-Two Sum 【Array】【Easy】
题目:
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.
输入一个数组和target,要在一个数组中找到两个数字,其和为target,从小到大输出数组中两个数字的位置。题目中假设有且仅有一个答案。
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
题目思路:
① 如果直接暴力解决,时间复杂度为(O(n^2)),两个循环,分别从数组中找到两个num,两者相加等于target。
②那么如果给定的数组是有序的会不会降低难度呢?如果是有序的数组,那么我们可以用“夹逼定理”来处理。简单来说就是首尾相加,如果比target大,则将尾数左移,如果小了首尾右移,直到两个数相加刚好等于target,那么我们可以先将数组排序,然后用双指针向中间“夹逼”,这种方法的时间复杂度为(O(nlogn)。这种方法要注意的是排序的时候要记录数组原来的位置,然后再排序。
③还有一种方法,使用字典。在python里面有一个dictionary的和C++ 的map功能一样。首先,我们建立一个字典,d = {},字典的key是数组的值num,value是相应的位置, 然后只要满足 num 和 target - num都在字典里面则找到答案。开始时字典是空的,从列表中开始读取值,当 num 和 target - num 都找到时,即停止寻找。这种方法的时间复杂度是(O(n)),不过空间复杂度是 O(MAXN)。
代码:
②
class Solution: # @return a tuple, (index1, index2) def twoSum(self, num, target): # sort sorted_num = sorted(num) # two points left = 0 right = len(num) - 1 res = [] while (left < right): sum = sorted_num[left] + sorted_num[right] if sum == target: # find out index break; elif sum > target: right -= 1 else: left += 1 if left == right: return -1, -1 else: pos1 = num.index(sorted_num[left]) + 1 pos2 = num.index(sorted_num[right]) + 1 if pos1 == pos2: # find again pos2 = num[pos1:].index(sorted_num[right]) + pos1 + 1 return min(pos1, pos2), max(pos1, pos2)
③
class Solution(object): def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ d = {}# d is a dictionary to map the value of nums and the index in nums size = 0 while size < len(nums): if not nums[size] in d: d[nums[size]] = size + 1 #if nums[size] doesn't exist in d ,create it if target - nums[size] in d: #if nums[size] and target - nums[size] are both in d if d[target-nums[size]] < size + 1: # one situation should be minded nums[size] == target - nums[size] ans = [d[target - nums[size]] , size + 1]# for example [0,1,2] 0 and [0,1,2,0],0 return ans size = size + 1
PS:
如果想要使用pycharm进行调试,可以使用主函数(放在程序最后):
if __name__ == "__main__": s = Solution() print(s.twoSum([4, 2, 3, 6], 8))
结果: