【问题描述】[中等]

编写一个函数来查找字符串数组中的最长公共前缀。如果不存在公共前缀，返回空字符串 ""。示例 1:输入: ["flower","flow","flight"]
输出: "fl"
示例 2:输入: ["dog","racecar","car"]
输出: ""
解释: 输入不存在公共前缀。
说明:所有输入只包含小写字母 a-z 。

【解答思路】

1. 横向扫描

时间复杂度：O(N^2) 空间复杂度：O(1)

class Solution {public String longestCommonPrefix(String[] strs) {if (strs == null || strs.length == 0) {return "";}String prefix = strs[0];int count = strs.length;for (int i = 1; i < count; i++) {prefix = longestCommonPrefix(prefix, strs[i]);if (prefix.length() == 0) {break;}}return prefix;}public String longestCommonPrefix(String str1, String str2) {int length = Math.min(str1.length(), str2.length());int index = 0;while (index < length && str1.charAt(index) == str2.charAt(index)) {index++;}return str1.substring(0, index);}
}

2. 纵向扫描

时间复杂度：O(N^2) 空间复杂度：O(1)

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

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

2. 二分法

时间复杂度：O(mnlogm) 空间复杂度：O(1)

class Solution {public String longestCommonPrefix(String[] strs) {if (strs == null || strs.length == 0) {return "";}int minLength = Integer.MAX_VALUE;for (String str : strs) {minLength = Math.min(minLength, str.length());}int low = 0, high = minLength;while (low < high) {int mid = (high - low + 1) / 2 + low;if (isCommonPrefix(strs, mid)) {low = mid;} else {high = mid - 1;}}return strs[0].substring(0, low);}public boolean isCommonPrefix(String[] strs, int length) {String str0 = strs[0].substring(0, length);int count = strs.length;for (int i = 1; i < count; i++) {String str = strs[i];for (int j = 0; j < length; j++) {if (str0.charAt(j) != str.charAt(j)) {return false;}}}return true;}
}

4. 分治

复杂度

class Solution {public String longestCommonPrefix(String[] strs) {if (strs == null || strs.length == 0) {return "";}int minLength = Integer.MAX_VALUE;for (String str : strs) {minLength = Math.min(minLength, str.length());}int low = 0, high = minLength;while (low < high) {int mid = (high - low + 1) / 2 + low;if (isCommonPrefix(strs, mid)) {low = mid;} else {high = mid - 1;}}return strs[0].substring(0, low);}public boolean isCommonPrefix(String[] strs, int length) {String str0 = strs[0].substring(0, length);int count = strs.length;for (int i = 1; i < count; i++) {String str = strs[i];for (int j = 0; j < length; j++) {if (str0.charAt(j) != str.charAt(j)) {return false;}}}return true;}
}

【总结】

1.纵横交错 二分分治

2. 字符串/数组题目遍历 暴力再优化

转载链接：https://leetcode-cn.com/problems/longest-common-prefix/solution/zui-chang-gong-gong-qian-zhui-by-leetcode-solution/