# 2259. Remove Digit From Number to Maximize Result

Easy
You are given a string `number` representing a positive integer and a character `digit`.
Return the resulting string after removing exactly one occurrence of `digit` from `number` such that the value of the resulting string in decimal form is maximized. The test cases are generated such that `digit` occurs at least once in `number`.
Example 1:
Input: number = "123", digit = "3"
Output:
"12"
Explanation:
There is only one '3' in "123". After removing '3', the result is "12".
Example 2:
Input: number = "1231", digit = "1"
Output:
"231"
Explanation:
We can remove the first '1' to get "231" or remove the second '1' to get "123".
Since 231 > 123, we return "231".
Example 3:
Input: number = "551", digit = "5"
Output:
"51"
Explanation:
We can remove either the first or second '5' from "551".
Both result in the string "51".
Constraints:
• `2 <= number.length <= 100`
• `number` consists of digits from `'1'` to `'9'`.
• `digit` is a digit from `'1'` to `'9'`.
• `digit` occurs at least once in `number`.

### 解題

func removeDigit(number string, digit byte) string {
ans := "0"
for i:=0; i<len(number); i++{
if number[i]==digit{
num := number[:i]+number[i+1:]
if len(num)>len(ans) {
ans = num
} else if len(num)==len(ans) {
for j:=0; j<len(num); j++ {
if num[j]>ans[j] {
ans = num
break
} else if num[j]<ans[j] {
break
}
}
}
}
}
return ans
}