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q0438.c
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q0438.c
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//
// q0438.c
// LeetCode
//
// Created by NowOrNever on 18/05/2020.
// Copyright © 2020 DoubleL. All rights reserved.
//
#include "q0438.h"
#include "common.h"
//438. Find All Anagrams in a String
//Medium
//
//2620
//
//166
//
//Add to List
//
//Share
//Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
//
//Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
//
//The order of output does not matter.
//
//Example 1:
//
//Input:
//s: "cbaebabacd" p: "abc"
//
//Output:
//[0, 6]
//
//Explanation:
//The substring with start index = 0 is "cba", which is an anagram of "abc".
//The substring with start index = 6 is "bac", which is an anagram of "abc".
//Example 2:
//
//Input:
//s: "abab" p: "ab"
//
//Output:
//[0, 1, 2]
//
//Explanation:
//The substring with start index = 0 is "ab", which is an anagram of "ab".
//The substring with start index = 1 is "ba", which is an anagram of "ab".
//The substring with start index = 2 is "ab", which is an anagram of "ab".
//Accepted
//217,677
//Submissions
//521,596
int* findAnagrams(char * s, char * p, int* returnSize){
*returnSize = 0;
size_t p_len = strlen(p);
size_t s_len = strlen(s);
int *result = malloc(sizeof(int) * (s_len + 1));
if (s_len <= 0 || s_len < p_len) {
return result;
}
int ascii[128] = {0};
// p的字符
size_t count = p_len;
while (*p) {
ascii[*p++]++;
}
// first window
char *start = s;
char *end = s;
for (int i = 0; i < p_len; i++) {
if(ascii[*end++]-- > 0) count--;
}
if (!count){
result[(*returnSize)++] = (int)(start - s);
}
// window move
while (*end) {
if(++ascii[*start++] > 0) count++;
if(ascii[*end++]-- > 0) count--;
if (!count) {
result[(*returnSize)++] = (int)(start - s);
}
}
return result;
}
int question438(void){
char *s1 =
"";
char *s2 =
"a";
int size = 0;
int *result = findAnagrams(s1, s2, &size);
for (int i = 0; i < size; i++) {
printf("%d ",result[i]);
}
return 0;
}