bluecode-字符替换与最长子串问题
问题描述
小R得到了一个由大写字母组成的字符串，长度为 n。她可以对字符串中的字符进行修改，每次操作允许将某个位置的字符修改为任意字符。例如，字符串 ABC 的第一个字符 A 改为 B，则字符串变为 BBC。她最多可以进行 k 次这样的修改。
小R想知道，通过最多 k 次修改后，字符串中由最多两种不同字母组成的最长连续子串的长度是多少。
测试样例
样例1：
输入：
n = 6 ,k = 1 ,inp = "ABCBAD"
输出：5
样例2：
输入：
n = 5 ,k = 1 ,inp = "AEABD"
输出：4
样例3：
输入：
n = 8 ,k = 2 ,inp = "AAAABBCD"
输出：8
#include <algorithm>
#include <iostream>
#include <string>
#include <unordered_set>
#include <vector>
using namespace std;
int solution(int n, int k, std::string s) {
if (n <= 2)
return n;
int max_len = 0;
unordered_set<char> char_set(s.begin(), s.end());
vector<char> chars;
for (auto &p : char_set)
chars.push_back(p);
for (char char1 : chars) {
for (char char2 : chars) {
if (char1 == char2)
continue;
int left = 0;
int count1 = 0;
int count2 = 0;
int total = 0;
for (int right = 0; right < n; ++right) {
total++;
if (s[right] == char1)
count1++;
else if (s[right] == char2)
count2++;
while (total - count1 - count2 > k) {
if (s[left] == char1)
count1--;
else if (s[left] == char2)
count2--;
left++;
total--;
}
max_len = max(max_len, total);
}
}
}
for (char char1 : chars) {
int left = 0;
int count = 0;
int total = 0;
for (int right = 0; right < n; ++right) {
total++;
if (s[right] == char1)
count++;
while (total - count > k) {
if (s[left] == char1)
count--;
++left;
--total;
}
max_len = max(max_len, total);
}
}
return max_len;
}
int main() {
std::cout << (solution(6, 1, "ABCBAD") == 5)
<< std::endl; // Expected: 1 (true)
std::cout << (solution(5, 1, "AEABD") == 4)
<< std::endl; // Expected: 1 (true)
std::cout << (solution(8, 2, "AAAABBCD") == 8)
<< std::endl; // Expected: 1 (true)
return 0;
}package main
import (
"fmt"
)
// solution 返回通过最多 k 次修改后，包含最多两种字母的最长子串长度
func solution(n, k int, s string) int {
if n <= 2 {
return n
}
maxLen := 0
// 获取唯一字符集合
charSet := make(map[rune]bool)
for _, c := range s {
charSet[c] = true
}
// 将字符放入切片
chars := make([]rune, 0, len(charSet))
for c := range charSet {
chars = append(chars, c)
}
// 枚举两种字母组合
for _, char1 := range chars {
for _, char2 := range chars {
if char1 == char2 {
continue
}
left, count1, count2, total := 0, 0, 0, 0
for right := 0; right < n; right++ {
total++
if rune(s[right]) == char1 {
count1++
} else if rune(s[right]) == char2 {
count2++
}
// 缩小窗口直到修改次数 <= k
for total-count1-count2 > k {
if rune(s[left]) == char1 {
count1--
} else if rune(s[left]) == char2 {
count2--
}
left++
total--
}
if total > maxLen {
maxLen = total
}
}
}
}
// 处理单一字母的情况
for _, char1 := range chars {
left, count, total := 0, 0, 0
for right := 0; right < n; right++ {
total++
if rune(s[right]) == char1 {
count++
}
for total-count > k {
if rune(s[left]) == char1 {
count--
}
left++
total--
}
if total > maxLen {
maxLen = total
}
}
}
return maxLen
}
func main() {
testCases := []struct {
n int
k int
s string
}{
{6, 1, "ABCBAD"},
{5, 1, "AEABD"},
{8, 2, "AAAABBCD"},
}
for _, tc := range testCases {
fmt.Println(solution(tc.n, tc.k, tc.s))
}
}def solution(n, k, s):
if n <= 2:
return n
max_len = 0
# 获取唯一字符集合
chars = set(s)
# 枚举两种字母组合
for char1 in chars:
for char2 in chars:
if char1 == char2:
continue
left = 0
count1 = 0 # char1 的数量
count2 = 0 # char2 的数量
total = 0 # 窗口总长度
for right in range(n):
total += 1
if s[right] == char1:
count1 += 1
elif s[right] == char2:
count2 += 1
# 缩小窗口直到修改次数 <= k
while total - count1 - count2 > k:
if s[left] == char1:
count1 -= 1
elif s[left] == char2:
count2 -= 1
left += 1
total -= 1
max_len = max(max_len, total)
# 处理单一字母的情况
for char1 in chars:
left = 0
count = 0
total = 0
for right in range(n):
total += 1
if s[right] == char1:
count += 1
while total - count > k:
if s[left] == char1:
count -= 1
left += 1
total -= 1
max_len = max(max_len, total)
return max_len
# 测试代码
if __name__ == "__main__":
test_cases = [
(6, 1, "ABCBAD"),
(5, 1, "AEABD"),
(8, 2, "AAAABBCD"),
]
for n, k, s in test_cases:
print(solution(n, k, s))