LintCode ---- 刷题总结

对于一个给定的 source 字符串和一个 target 字符串，你应该在 source 字符串中找出 target 字符串出现的第一个位置(从0开始)。如果不存在，则返回 -1。

基本：两重for循环，时间复杂度O(n^2)。

 class Solution {
     /**
      * Returns a index to the first occurrence of target in source,
      * or -1  if target is not part of source.
      * @param source string to be scanned.
      * @param target string containing the sequence of characters to match.
      */
     public int strStr(String source, String target) {
         //write your code here
         if (source == null || target == null) {
             return -1;
         }
         for (int i = 0; i < source.length() - target.length() + 1; i++) {
             int j = 0;
             for (j = 0; j < target.length(); j++) {
                 if (source.charAt(i + j) != target.charAt(j)) {
                     break;
                 }
             }
             if (j == target.length()) {
                 return i;
             }
         }
         return -1;
     }
 }

高级：Rabin Karp算法，时间复杂度为O(n+m)，n为源字符串长度，m为目标字符串长度。该算法时间复杂度与KMP算法一样，但是比KMP简单，不建议使用KMP，不仅写起来麻烦而且容易错。

 public class Solution {
     private static int BASE = 1000000;
     /**
      * @param source a source string
      * @param target a target string
      * @return an integer as index
      */
     public int strStr2(String source, String target) {
         // Write your code here
         if (source == null || target == null) {
             return -1;
         }
         int m = target.length();
         if (m == 0) {
             return 0;
         }
         //31 ^ m
         int power = 1;
         for (int i = 0; i < m; i++) {
             power = (power * 31) % BASE;
         }
         //targetCode
         int targetCode = 0;
         for (int i = 0; i < m; i++) {
             targetCode = (targetCode * 31 + target.charAt(i)) % BASE;
         }
         //hashCode
         int hashCode = 0;
         for (int i = 0; i < source.length(); i++) {
             hashCode = (hashCode * 31 + source.charAt(i)) % BASE;
             if (i < m - 1) {
                 continue;
             }
             if (i >= m) {
                 hashCode = hashCode - (source.charAt(i - m) * power) % BASE;
                 if (hashCode < 0) {
                     hashCode += BASE;
                 }
             }
             if (targetCode == hashCode) {
                 if (source.substring(i - m + 1, i + 1).equals(target)) {
                     return i - m + 1;
                 }
             }
         }
         return -1;
     }
 }