LeetCode1两数之和

题目

给定一个整数数组和一个目标值，找出数组中和为目标值的两个数。你可以假设每个输入只对应一种答案，且同样的元素不能被重复利用。

示例:

给定 nums = [2, 7, 11, 15], target = 9

因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

方法一：

双层循环

class Solution(object):
    def twoSum(self,nums,target):
        flags = [0,0,0,0]
        l = len(nums)
        for a in range(l):
            flags[a] = 1
            for b in range(l):
                if flags[b] == 0:
                    if a != b:
                        if nums[a] + nums[b] == target:
                            return [a,b]

nums = [2,7,11,15]
target = 9
test_o = Solution()
result = test_o.twoSum(nums,target)
print(result)

flag是去重复运算 a+b和b+a是一模一样的

方法二

一层循环

class Solution1(object):
    def towSum1(self,nums1,target1):
        l = len(nums1)
        for a in range(l):
            one_num = nums1[a]
            otherOne = target1 - one_num
            if otherOne in nums1:
                b = nums1.index(otherOne)
                if a != b:
                    if a>b:
                        return [b,a]
                    return [a,b]

nums1 = [2,7,11,15]
target1 = 9
xxx = Solution1()
result1 = xxx.towSum1(nums1,target1)
print(result1)

方法三

一层循环优化

class Solution3(object):
    def towSum2(self,nums2,target2):
        l = len(nums2)
        dict_nums = {nums2[i]:i for i in range(l)}
        for a in range(l):
            one_num = nums2[a]
            other_one = target2 - one_num
            if other_one in dict_nums and a != dict_nums[other_one]:
                return [a,dict_nums[other_one]]

nums2 = [2,7,11,15]
target2 = 9
yyy = Solution3()
result2 = yyy.towSum2(nums2,target2)
print(result2)

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2020年1月4日于长沙