0028. Implement strStr() (E)

Implement strStr() (E)

题目

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().

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题意

找到匹配的字串。

思路

暴力法\(O(MN)\)。

KMP。

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代码实现

Java

暴力

class Solution {
    public int strStr(String haystack, String needle) {
        if (needle.equals("")) {
            return 0;
        }

        for (int i = 0; i < haystack.length(); i++) {
            if (haystack.length() - i < needle.length()) {
                break;
            }
            if (haystack.charAt(i) == needle.charAt(0)) {
                int j = i, k = 0;
                while (k < needle.length()) {
                    if (haystack.charAt(j) != needle.charAt(k)) {
                        break;
                    }
                    j++;
                    k++;
                }
                if (k == needle.length()) {
                    return i;
                }
            }
        }

        return -1;
    }
}

KMP

class Solution {
    public int strStr(String haystack, String needle) {
        if (needle.length() == 0) {
            return 0;
        }
        
        int[] next = new int[needle.length()];
        int i = 1, p = 0;
        while (i < needle.length()) {
            if (needle.charAt(i) == needle.charAt(p)) {
                next[i++] = ++p;
            } else if (p > 0) {
                p = next[p - 1];
            } else {
                next[i++] = 0;
            }
        }
        
        i = 0;
        p = 0;
        while (i < haystack.length()) {
            if (haystack.charAt(i) == needle.charAt(p)) {
                i++;
                p++;
                if (p == needle.length()) {
                    return i - p;
                }
            } else if (p > 0) {
                p = next[p - 1];
            } else {
                i++;
            }
        }
        
        return -1;
    }
}

JavaScript

暴力

/**
 * @param {string} haystack
 * @param {string} needle
 * @return {number}
 */
var strStr = function (haystack, needle) {
  if (needle.length === 0) {
    return 0
  }

  for (let i = 0; i < haystack.length; i++) {
    if (haystack.length - i < needle.length) {
      break
    }
    let isMatch = true
    for (let j = 0; j < needle.length; j++) {
      if (needle[j] !== haystack[i + j]) {
        isMatch = false
        break
      }
    }
    if (isMatch) {
      return i
    }
  }
  return -1
}

KMP

/**
 * @param {string} haystack
 * @param {string} needle
 * @return {number}
 */
var strStr = function (haystack, needle) {
  if (needle.length === 0) {
    return 0
  }
  
  let next = new Array(needle.length).fill(0)
  let p = 0, q = 1
  while (q < needle.length) {
    if (needle[q] === needle[p]) {
      next[q++] = ++p
    } else if (p > 0) {
      p = next[p - 1]
    } else {
      next[q++] = 0
    }
  }

  p = 0, q = 0
  while (q < haystack.length) {
    if (needle[p] === haystack[q]) {
      p++
      q++
      if (p === needle.length) {
        return q - needle.length
      }
    } else if (p > 0) {
      p = next[p - 1]
    } else {
      q++
    }
  }
  
  return -1
}