# 2566. Maximum Difference by Remapping a Digit
###### tags: `Leetcode` `Easy`
Link: https://leetcode.com/problems/maximum-difference-by-remapping-a-digit/description/
## 思路
### 思路一 $O(n)$ $O(n)$
找到第一个非9的数字 我们要将它全部替换成9 就会得到maxValue
找到第一个非0的数字 将它全部替换成0 就会得到minValue
### 思路二 $O(N)$ $O(1)$
尝试把所有数字0...9挨个换成0或9 找出他们中的最大最小值 求差就是答案
## Code
### 思路一

class Solution {
    public int minMaxDifference(int num) {
        char[] val = String.valueOf(num).toCharArray();
        char[] temp = String.valueOf(num).toCharArray();
        //max
        char remap = '9';
        for(int i=0; i<val.length; i++){
            if(val[i]!='9'){
                remap = val[i];
                break;
            }
        }
        for(int i=0; i<val.length; i++){
            if(temp[i]==remap){
                temp[i] = '9';
            }
        }
        int max = Integer.valueOf(String.valueOf(temp));
        //min
        remap = '0';
        for(int i=0; i<val.length; i++){
            if(val[i]!='0'){
                remap = val[i];
                break;
            }
        }
        for(int i=0; i<val.length; i++){
            if(val[i]==remap){
                val[i] = '0';
            }
        }
        int min = Integer.valueOf(String.valueOf(val));
        return max-min;
    }
}

### 思路二

class Solution {
    public int minMaxDifference(int num) {
        int maxVal = num, minVal = num;
        for(int i=0; i<=9; i++){
            int hi = 0, lo = 0, mul = 1, n = num;
            while(n>0){
                boolean replace = (n%10==i);
                hi += mul*(replace?9:n%10);
                lo += mul*(replace?0:n%10);
                n /= 10;
                mul *= 10;
            }
            maxVal = Math.max(hi, maxVal);
            minVal = Math.min(lo, minVal);
        }
        return maxVal-minVal;
    }
}