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题目
Implement strStr().
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Clarification:
What should we return when needle is an empty string? This is a great question to ask during an interview.
For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s strstr() and Java’s indexOf().
Example 1:
Input: haystack = "hello", needle = "ll"
Output: 2
Example 2:
Input: haystack = "aaaaa", needle = "bba"
Output: -1
Example 3:
Input: haystack = "", needle = ""
Output: 0
Constraints:
0 <= haystack.length, needle.length <= 5 * 104
haystack and needle consist of only lower-case English characters.
1. 穷举解法
两个for循环匹配即可，这里容易遗漏的点是判断needle是否为空字符串"", 是的话返回0.
class Solution {
public int strStr(String haystack, String needle) {
// check null
if (haystack == null && needle == null) {
return 0;
}
if (haystack == null || needle == null) {
return -1;
}
int hlen = haystack.length();
int nlen = needle.length();
// check ""
if (nlen == 0) {
return 0;
}
for(int i = 0; i <= hlen - nlen; i++) {
for(int k = 0; k < nlen; k++) {
if (haystack.charAt(i + k) != needle.charAt(k)) {
break;
}
if (k == nlen - 1) {
return i;
}
}
}
return -1;
}
}
2. 优雅的写法
假设两个字符串都不为空，可以用下面优雅的写法。
class Solution {
public int strStr(String haystack, String needle) {
for (int i = 0; ; i++) {
for (int k = 0; ; k++) {
if (k == needle.length()) return i;
if (i + k == haystack.length()) return -1;
if (haystack.charAt(i + k) != needle.charAt(k)) break;
}
}
}
}