实现 strStr()
约 429 字大约 1 分钟
实现 strStr()
写法
class Solution {
/**
* 基于窗口滑动的算法
* <p>
* 时间复杂度:O(m*n)
* 空间复杂度:O(1)
* 注:n为haystack的长度,m为needle的长度
*/
public int strStr(String haystack, String needle) {
int m = needle.length();
// 当 needle 是空字符串时我们应当返回 0
if (m == 0) {
return 0;
}
int n = haystack.length();
if (n < m) {
return -1;
}
int i = 0;
int j = 0;
while (i < n - m + 1) {
// 找到首字母相等
while (i < n && haystack.charAt(i) != needle.charAt(j)) {
i++;
}
if (i == n) {// 没有首字母相等的
return -1;
}
// 遍历后续字符,判断是否相等
i++;
j++;
while (i < n && j < m && haystack.charAt(i) == needle.charAt(j)) {
i++;
j++;
}
if (j == m) {// 找到
return i - j;
} else {// 未找到
i -= j - 1;
j = 0;
}
}
return -1;
}
}
// 方法一
class Solution {
public void getNext(int[] next, String s){
int j = -1;
next[0] = j;
for (int i = 1; i < s.length(); i++){
while(j >= 0 && s.charAt(i) != s.charAt(j+1)){
j=next[j];
}
if(s.charAt(i) == s.charAt(j+1)){
j++;
}
next[i] = j;
}
}
public int strStr(String haystack, String needle) {
if(needle.length()==0){
return 0;
}
int[] next = new int[needle.length()];
getNext(next, needle);
int j = -1;
for(int i = 0; i < haystack.length(); i++){
while(j>=0 && haystack.charAt(i) != needle.charAt(j+1)){
j = next[j];
}
if(haystack.charAt(i) == needle.charAt(j+1)){
j++;
}
if(j == needle.length()-1){
return (i-needle.length()+1);
}
}
return -1;
}
}
class Solution {
//前缀表(不减一)Java实现
public int strStr(String haystack, String needle) {
if (needle.length() == 0) return 0;
int[] next = new int[needle.length()];
getNext(next, needle);
int j = 0;
for (int i = 0; i < haystack.length(); i++) {
while (j > 0 && needle.charAt(j) != haystack.charAt(i))
j = next[j - 1];
if (needle.charAt(j) == haystack.charAt(i))
j++;
if (j == needle.length())
return i - needle.length() + 1;
}
return -1;
}
private void getNext(int[] next, String s) {
int j = 0;
next[0] = 0;
for (int i = 1; i < s.length(); i++) {
while (j > 0 && s.charAt(j) != s.charAt(i))
j = next[j - 1];
if (s.charAt(j) == s.charAt(i))
j++;
next[i] = j;
}
}
}