Implement strStr().
Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
最直接的方式: 从头开始逐渐匹配,若不一样,则后移一位继续比较.
时间复杂度:O(M*N) M/N分别为两个字符串的长度
public int strStr(String haystack, String needle) {
if(needle==null||haystack==null)
return -1;
int len1=haystack.length();
int len2=needle.length();
if(len2>len1)
return -1;
int j;
for(int i=0;i<len1-len2+1;i++){
for(j=0;j<len2;j++){
if(haystack.charAt(i+j)!=needle.charAt(j))
break;
}
if(j==len2)
return i;
}
return -1;
}
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