1.LeetCode之Python版本的“两数之和”(简单题)

顺序遍历数组，每访问一个数，就用目标数减去这个数，查看减法的结果是否存在于数组中。为了避免返回的结果是自己和自己相加，首先将自己先移除，然后再判断。如果在的话，就输出两个数的下标；如果不在的话，就将刚才移掉的自己再插入到原来的位置上保持不变，访问下一个数。

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            target1 = nums[i]
            target2 = target-target1
            nums.remove(target1)
            if target2 in nums:
                return [i,nums.index(target2)+1]
            else:
                nums.insert(i,target1)

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            target1 = nums[i]
            target2 = target-target1
            nums.remove(target1)
            if target2 in nums:
                return [i,nums.index(target2)+1]
            else:
                nums.insert(i,target1)
solution = Solution()
print(solution.twoSum([2,7,11,15], 9))

拒绝内耗，永远心存善念。