Fix outdate number error of findRepeatNumber

[tczhangzhi]

原方法忘记在交换后更新当前i指针对应的num，导致测试用例 nums = [1, 3, 4, 0, 2, 5, 3] 返回1（应该返回3）:

from typing import List

def findRepeatNumber(nums: List[int]) -> int:
    for i, num in enumerate(nums):
        while i != num:
            if num == nums[num]:
                return num
            nums[i], nums[num] = nums[num], nums[i]
           # 原方法缺少这一行：
           # num = nums[i]
    return -1

nums = [1, 3, 4, 0, 2, 5, 3]
findRepeatNumber(nums)

因此，本次提交作出了修正：

def findRepeatNumber(nums: List[int]) -> int:
    for i, num in enumerate(nums):
        while i != num:
            if num == nums[num]:
                return num
            nums[i], nums[num] = nums[num], nums[i]
            num = nums[i]
    return -1

修正后可通过三种边界测试和其他笔者想到的任何符合题目要求情况：

nums = [1, 3, 4, 0, 2, 5, 3] # should return 3
nums = [2, 4, 1, 0, 2, 5, 3] # should return 2
nums = [2, 4, 1, 0, 6, 5, 3] # should return -1
findRepeatNumber(nums)

另，虽然复杂度相同，但读者更偏爱只使用一个循环表示该过程：

def findRepeatNumber(nums: List[int]) -> int:
    i = 0
    while i < len(nums):
        print(nums)
        if nums[i] == i:
            i += 1
            continue
        if nums[nums[i]] == nums[i]: return nums[i]
        nums[nums[i]], nums[i] = nums[i], nums[nums[i]]
    return -1

nums = [1, 3, 4, 0, 2, 5, 3]
findRepeatNumber(nums)