[LeetCode] 28. Find the Index of the First Occurrence in a String
grandyang opened this issue · comments
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Given two strings needle
and haystack
, return the index of the first occurrence of needle
in haystack
, or -1
if needle
is not part of haystack
.
Example 1:
**Input:** haystack = "sadbutsad", needle = "sad"
**Output:** 0
**Explanation:** "sad" occurs at index 0 and 6.
The first occurrence is at index 0, so we return 0.
Example 2:
**Input:** haystack = "leetcode", needle = "leeto"
**Output:** -1
**Explanation:** "leeto" did not occur in "leetcode", so we return -1.
Constraints:
1 <= haystack.length, needle.length <= 104
haystack
andneedle
consist of only lowercase English characters.
这道题让在一个字符串中找另一个字符串第一次出现的位置,那首先要做一些判断,如果子字符串为空,则返回0,如果子字符串长度大于母字符串长度,则返回 -1。然后开始遍历母字符串,这里并不需要遍历整个母字符串,而是遍历到剩下的长度和子字符串相等的位置即可,这样可以提高运算效率。然后对于每一个字符,都遍历一遍子字符串,一个一个字符的对应比较,如果对应位置有不等的,则跳出循环,如果一直都没有跳出循环,则说明子字符串出现了,则返回起始位置即可,代码如下:
class Solution {
public:
int strStr(string haystack, string needle) {
if (needle.empty()) return 0;
int m = haystack.size(), n = needle.size();
if (m < n) return -1;
for (int i = 0; i <= m - n; ++i) {
int j = 0;
for (j = 0; j < n; ++j) {
if (haystack[i + j] != needle[j]) break;
}
if (j == n) return i;
}
return -1;
}
};
我们也可以写的更加简洁一些,开头直接套两个 for 循环,不写终止条件,然后判断假如j到达 needle 的末尾了,此时返回i;若此时 i+j 到达 haystack 的长度了,返回 -1;否则若当前对应的字符不匹配,直接跳出当前循环,参见代码如下:
解法二:
class Solution {
public:
int strStr(string haystack, string needle) {
for (int i = 0; ; ++i) {
for (int j = 0; ; ++j) {
if (j == needle.size()) return i;
if (i + j == haystack.size()) return -1;
if (needle[j] != haystack[i + j]) break;
}
}
return -1;
}
};
Github 同步地址:
类似题目:
参考资料:
https://leetcode.com/problems/implement-strstr/
https://leetcode.com/problems/implement-strstr/discuss/12807/Elegant-Java-solution
https://leetcode.com/problems/implement-strstr/discuss/12956/C%2B%2B-Brute-Force-and-KMP
LeetCode All in One 题目讲解汇总(持续更新中...)
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为什么不用一下kmp呢?
class Solution {
public:
int strStr(string s, string p) {
if(p.empty()) return 0;
vector<int> next_char_to_be_matched(p.size(), 0);
for(int i = 1, k= 0; i < p.size(); ++i){
while(k && p[i] != p[k]) k = next_char_to_be_matched[k-1];
if(p[i] == p[k] ) ++k;
next_char_to_be_matched[i] = k;
}
for(int i = 0, k= 0; i < s.size(); ++i){
while(k && s[i] != p[k]) k = next_char_to_be_matched[k-1];
if(s[i] == p[k] ) ++k;
if(k == p.size()) return i+1-k;
}
return -1;
}
};