【LeetCode】14. Longest Common Prefix

Problem

【LeetCode】14. Longest Common Prefix

Algorithmic thinking

方法一:水平扫描法
【LeetCode】14. Longest Common Prefix
【LeetCode】14. Longest Common Prefix

Java解法:

 public String longestCommonPrefix(String[] strs) {
    if (strs.length == 0) return "";
    String prefix = strs[0];
    for (int i = 1; i < strs.length; i++)
        while (strs[i].indexOf(prefix) != 0) {
            prefix = prefix.substring(0, prefix.length() - 1);
            if (prefix.isEmpty()) return "";
        }        
    return prefix;
}
 

【LeetCode】14. Longest Common Prefix

算法二:水平扫描
算法

想象数组的末尾有一个非常短的字符串, 使用上述方法依旧会进行 S​S​ 次比较。 优化这类情况的一种方法就是水平扫描。 我们从前往后枚举字符串的每一列,先比较每个字符串相同列上的字符(即不同字符串相同下标的字符)然后再进行对下一列的比较。

Java 解法

public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";
    for (int i = 0; i < strs[0].length() ; i++){
        char c = strs[0].charAt(i);
        for (int j = 1; j < strs.length; j ++) {
            if (i == strs[j].length() || strs[j].charAt(i) != c)
                return strs[0].substring(0, i);             
        }
    }
    return strs[0];
}

【LeetCode】14. Longest Common Prefix

相关文档(力扣):
https://leetcode-cn.com/problems/longest-common-prefix/solution/


Python3 solution

class Solution:
    # @return a string
    def longestCommonPrefix(self, strs):
        if not strs:
            return ""

        for i, letter_group in enumerate(zip(*strs)):
            if len(set(letter_group)) > 1:
                return strs[0][:i]
        else:
            return min(strs)


if __name__ == '__main__':
    sl_1 = ['flow', 'fawer', 'flower']
    result = Solution().longestCommonPrefix(sl_1)
    print(result)