LeetCode - 28. Implement strStr()
2017-02-19 22:19
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题目:
Implement strStr().Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
思路与步骤:
两个字符串逐个字符比较,如果两个字符串第一个不同,则大的下标加1,小的不变,依次循环;如果相同,则依次逐个比较后面的,全部相同则返回;如果不是全部相同,则小的下标重新设为0,大的在最开始的基础上前进1。
程序通过后,经过几次简化,最后写出了比较简洁的程序。
编程实现:
C程序:
int strStr(char* haystack, char* needle) { int hslen = strlen(haystack), nlen = strlen(needle); if(nlen == 0) return 0; if(nlen > hslen) return -1; int i=0, j=0; while (i <= hslen - nlen){ if(haystack[i] != needle[j]) i++; else{ while(haystack[i] == needle[j] && j < nlen-1){ i++; j++; } if(j == nlen-1 && haystack[i] == needle[j]) return i-j; else{ i = i-j+1; j = 0; } } } return -1; }
Java程序:
public class Solution { public int strStr(String haystack, String needle) { if(needle.length() == 0) return 0; if(needle.length() > haystack.length()) return -1; char[] hsChar = haystack.toCharArray(); char[] nChar = needle.toCharArray(); int i=0, j=0; while (i <= haystack.length()-needle.length()){ if((hsChar[i] != nChar[j])) i++; else{ while(hsChar[i] == nChar[j] && j<needle.length()-1){ i++; j++; } if(j==needle.length()-1 && hsChar[i] == nChar[j]) return i-j; else{ i = i-j+1; //注意这里 j = 0; } } } return -1; } }别人的简洁的算法(还没看)效率并没有很高:
public class Solution { public int strStr(String haystack, String needle) { //别人的简洁方法,未看 for (int i = 0; ; i++) { for (int j = 0; ; j++) { if (j == needle.length()) return i; if (i + j == haystack.length()) return -1; if (needle.charAt(j) != haystack.charAt(i + j)) break; } } } }
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