Description

You are given an integer num. You will apply the following steps exactly two times:

  • Pick a digit x (0 <= x <= 9).
  • Pick another digit y (0 <= y <= 9). The digit y can be equal to x.
  • Replace all the occurrences of x in the decimal representation of num by y.
  • The new integer cannot have any leading zeros, also the new integer cannot be 0.

Let a and b be the results of applying the operations to num the first and second times, respectively.

Return the max difference between a and b.

Example 1:

Input: num = 555
Output: 888
Explanation: The first time pick x = 5 and y = 9 and store the new integer in a.
The second time pick x = 5 and y = 1 and store the new integer in b.
We have now a = 999 and b = 111 and max difference = 888

Example 2:

Input: num = 9
Output: 8
Explanation: The first time pick x = 9 and y = 9 and store the new integer in a.
The second time pick x = 9 and y = 1 and store the new integer in b.
We have now a = 9 and b = 1 and max difference = 8

Example 3:

Input: num = 123456
Output: 820000

Example 4:

Input: num = 10000
Output: 80000

Example 5:

Input: num = 9288
Output: 8700

Constraints:

  • 1 <= num <= 10^8

分析

题目的意思是：给定一个数字，然后替换其中的数字得道x，然后再替换该数字得到要，求x和y的最大差，替换后第一位不能为0，且整个数字不能为0.我的想法很简单，把该数字变成字符串，替换其中的字符为9，遍历求出最大值。至于最小值，要考虑到替换的时候首字母为0的情况，如果出现这种情况，应该替换为1；排除首字母为0的情况，其他情况都可以替换成1，然后求出最小值。最后得到结果。

我看了一下其他人的解法，对与最大的数，从左向右遍历找到第一个不是9的数，并进行替换就行了；对于最小的数，如果首字母不为1，则把首字母替换为1就是最小了；如果首字母为1，则进行遍历，找到第一个不是01两字符的数字替换就行了。感觉比我的思路简单，哈哈哈。

代码

class Solution:
    def maxDiff(self, num: int) -> int:
        num=str(num)
        max_val=int(num)
        min_val=int(num)
        for i,ch in enumerate(num):
            t1=num
            t1=int(t1.replace(ch,'9'))
            max_val=max(max_val,t1)
            t2=num
            if(i!=0):
                t3=t2.replace(ch,'0')
                if(t3[0]=='0'):
                    t3=t2.replace(ch,'1')
                min_val=min(min_val,int(t3))
            elif(i==0):
                t2=t2.replace(ch,'1')
                min_val=min(min_val,int(t2))
        return max_val-min_val

参考文献

​​[LeetCode] [Python] Clean greedy solution with explanation​​