1375. 至少K个不同字符的子串

给定一个仅包含小写字母的字符串 S.

返回 S 中至少包含 k 个不同字符的子串的数量.

样例

样例 1:

输入: S = "abcabcabca", k = 4

输出: 0

解释: 字符串中一共就只有 3 个不同的字符.

样例 2:

输入: S = "abcabcabcabc", k = 3

输出: 55

解释: 任意长度不小于 3 的子串都含有 a, b, c 这三个字符.

比如,长度为 3 的子串共有 10 个, "abc", "bca", "cab" ... "abc"

长度为 4 的子串共有 9 个, "abca", "bcab", "cabc" ... "cabc"

...

长度为 12 的子串有 1 个, 就是 S 本身.

所以答案是 1 + 2 + ... + 10 = 55.

注意事项

10 ≤ length(S) ≤ 1,000,000
1 ≤ k ≤ 26

public class Solution {

/**

* @param s: a string

* @param k: an integer

* @return: the number of substrings there are that contain at least k distinct characters

public long kDistinctCharacters(String s, int k) {

int[] map = new int[126];

int size = 0;

long sum = 0;

int i = 0;

int len = s.length();

for (; i < len; i++) {

int c = s.charAt(i);

if (map[c] == 0) {

size++;

if (size == k) {

sum += len - i;

map[c]++;

break;

}

map[c]++;

}

for (int j = 0; j < len; j++) {

int c = s.charAt(j);

map[c]--;

if (map[c] > 0) {

sum += len - i;

} else {

size--;

for (i++; i < len; i++) {

c = s.charAt(i);

if (map[c] == 0) {

size++;

if (size == k) {

sum += len - i;

map[c]++;

break;

}

map[c]++;

}

if (i == len) {

break;

}

return sum;

}

public class Solution {

/**

* @param s: a string

* @param k: an integer

* @return: the number of substrings there are that contain at least k distinct characters

public long kDistinctCharacters(String s, int k) {

HashMap<Character, Integer> map = new HashMap<>(26);

long sum = 0;

int i = 0;

for (; i < s.length(); i++) {

char c = s.charAt(i);

if (map.containsKey(c)) {

int num = map.get(c);

num++;

map.put(c, num);

} else {

map.put(c, 1);

if (map.size() == k) {

sum += s.length() - i;

break;

}

for (int j = 0; j < s.length(); j++) {

char c = s.charAt(j);

int num = map.get(c);

num--;

if (num > 0) {

sum += s.length() - i;

map.put(c, num);

} else {

map.remove(c);

boolean has=true;

for (i++; i < s.length(); i++) {

c = s.charAt(i);

if (map.containsKey(c)) {

int size = map.get(c);

size++;

map.put(c, size);

} else {

map.put(c, 1);

if (map.size() == k) {

sum += s.length() - i;

has=false;

break;

}

if (has){

break;

}

return sum;

}