QUESTION:
Given two strings s and part, perform the following operation on s until all occurrences of the substring part are removed:
Find the leftmost occurrence of the substring part and remove it from s. Return s after removing all occurrences of part.
A substring is a contiguous sequence of characters in a string.
Example 1:
Input: s = "daabcbaabcbc", part = "abc"
Output: "dab"
Explanation: The following operations are done:
- s = "daabcbaabcbc", remove "abc" starting at index 2, so s = "dabaabcbc".
- s = "dabaabcbc", remove "abc" starting at index 4, so s = "dababc".
- s = "dababc", remove "abc" starting at index 3, so s = "dab".
Now s has no occurrences of "abc".
Example 2:
Input: s = "axxxxyyyyb", part = "xy"
Output: "ab"
Explanation: The following operations are done:
- s = "axxxxyyyyb", remove "xy" starting at index 4 so s = "axxxyyyb".
- s = "axxxyyyb", remove "xy" starting at index 3 so s = "axxyyb".
- s = "axxyyb", remove "xy" starting at index 2 so s = "axyb".
- s = "axyb", remove "xy" starting at index 1 so s = "ab".
Now s has no occurrences of "xy".
Constraints:
1 <= s.length <= 1000
1 <= part.length <= 1000
s and part consists of lowercase English letters.
EXPLANATION:
使用递归的方式, 每次都将对应的part删除. 直到没有part为止.
SOLUTION:
class Solution {
func removeOccurrences(_ s: String, _ part: String) -> String {
guard let range = s.range(of: part) else {
return s
}
return removeOccurrences(s.substring(to: range.lowerBound)+s.substring(from: range.upperBound), part)
}
}