LeetCode- twoSum-python

Description

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].

Solution

时间复杂度：O(n^2)

空间复杂度：O(1)

核心：
第i个元素分别之后的每一个元素（i+1, i+2...., i+n）求和，然后做判断

def twoSum(self, nums, target):
    for i in xrange(len(nums)):
        for j in xrange(i+1, len(nums)):
            if (nums[i] + nums[j] == target):
                return i, j

时间复杂度：O(n^2)

空间复杂度：O(n)

核心：

1. 对数组做排序
2. 较小的元素i（取第一个元素）和元素j（取最后一个元素），求和
3. 如何和等于目标值，返回元素的索引即可；如果和小于目标值，那么将元素i的索引加1；反之，那么将元素j的索引减1。
4. 再求和，进入步骤3。

def twoSum(self, nums, target):
    index = []
    numtosort = nums[:]
    numtosort.sort()
    i = 0; j = len(numtosort) - 1
    while i < j:
        if numtosort[i] + numtosort[j] == target:
            index.append(nums.index(numtosort[i]))
            index.append(len(nums) - nums[::-1].index(numtosort[j]) - 1)
            index.sort()
            break
        elif numtosort[i] + numtosort[j] < target:
            i = i + 1
        elif numtosort[i] + numtosort[j] > target:
            j = j - 1
    return (index[0],index[1])

时间复杂度： O(n)

空间复杂度： O(n)

核心：

1. 取第i个元素，计算目标值减去该元素值
2. 判断是否在字典中，如果在那么取字典中该元素的索引即可
3. 否则将元素索引及值存入字典，进入步骤1

def twoSum(self, nums, target):
        dict = {}
        for i in xrange(len(nums)):
            x = nums[i]
            if target-x in dict:
                return (dict[target-x], i)
            dict[x] = i

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