一、字符串问题

 

题目：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。

分析：命名有意义；正常的for循环即可；一定有非空检查；数组不要越界

1）常规算法：

2,not ac :循环条件不对，j在循环内定义，判断i+ j,返回值cuo ,

robinhap not ac ：target长度没判断，31 ^,长度选错，循环变量错，targethash算错，targetHash = targetHash * 31 % BASE + target.charAt(i) % BASE;没double check

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//非空检查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;}}// loop overif (j == target.length()) {return i;}}return -1;} } View Code

 

2）Rabin-Karp算法：利用hash表的原理，把字符串转为数字。最后通过比较hash值是否相同

思路：
base
1、异常检测
2、target长度，为空检测
3、计算31^m（power）：边乘边模
4、targethash: 边乘边加边模
5、hashcode :边乘边加边模
i < mi-1:继续
i>m-1:abcd - a ,负数检测
hashcode == targethash
double check

not ac : abcd -a ，hashCode = hashCode - source.charAt(i - m) * power % BASE;忘记模；判断hashcode 是否小于0，写在外面了

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 BASE = 1000000;public int strStr(String source, String target) {// write your code here//非空检查if (source == null || target == null) {return -1;}int m = target.length();//target长度检查if (m <= 0) {return 0;}//31^mint pow = 1;for (int i = 0; i < m; i++) {pow = pow * 31 % BASE;}//targetcodeint targetCode = 0;for (int i = 0; i < m; i++) {targetCode = (targetCode * 31 + target.charAt(i)) % BASE;}//hashCodeint hashCode = 0;for (int i = 0; i < source.length(); i++) {hashCode = (hashCode * 31 + source.charAt(i)) % BASE;if (i < m - 1) {continue;}// i// abcd - aif (i >= m) {hashCode = hashCode - (pow * source.charAt(i - m)) % BASE;}// hashcode < 0 ?if (hashCode < 0) {hashCode = hashCode + BASE;}//hash值是否相等if (hashCode == targetCode) {//double checkif (source.substring(i - m + 1, i + 1).equals(target)) {return i - m + 1;}}}return -1;} } View Code

 

转载于:https://www.cnblogs.com/lingli-meng/p/6511499.html

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