28. Implement strStr()

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = “hello”, needle = “ll”
Output: 2

Example 2:

Input: haystack = “aaaaa”, needle = “bba”
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s strstr() and Java’s indexOf()).

Solution1:

two pointers, 放在while循环中

// Time:O(n*m), Space:O(1)
class Solution {
 public:
  int strStr(string haystack, string needle) {
    int n = haystack.length(), m = needle.length();
    if (!m) return 0;

    int i = 0, j = 0;
    while (i < n) {
      if (haystack[i] == needle[j]) {
        int ok = 1, k = i;
        while (k < n && j < m) {
          if (haystack[k++] != needle[j++]) {
            ok = 0; break;
          }
        }
        if (ok && j == m && k <= n)
          return i;
        j = 0;
      }
      i++;
    }
    return -1;
  }

};

Solution2:

for循环, j作为needle的下标, i作为判断的头指针, i+j是haystack的下标,

1, 若两个指针指向的字符不同, 推出内层循环；

2, 如果j能够在上述条件下, 到达哨兵位置, 表明完全匹配;

3, 如果i+j到了haystack的哨兵位置, 扫描结束, 未找到;

// Time:O(n*m), Space:O(1)
class Solution {
 public:
  int strStr(string haystack, string needle) {
    for (int i = 0; ; i++) {
      for (int j = 0; ; j++) {
        if (j == needle.length()) return i;
        if (i+j == haystack.length()) return -1;
        if (haystack[i+j] != needle[j]) break;
      }
    }
  }
};

根据这个思路, 对Solution1进行简化,

用达到哨兵位置取代对ok变量的操作;

用i+j取代新增的k变量;

i的范围在n-m之内.

// 第一个版本的简化版本
// Time:O(n*m), Space:O(1)
class Solution {
 public:
  int strStr(string haystack, string needle) {
    int n = haystack.length(), m = needle.length();
    if (!m) return 0;

    int i = 0, j = 0;
    while (i <= n-m) {
      if (haystack[i] == needle[j]) {
        while (j < m && haystack[i+j] == needle[j]) j++;
        if (j == m) return i;
        j = 0;
      }
      i++;
    }
    return -1;
  }
};