leetcode 14. 最长公共前缀

直接法

直接法又分为竖向扫描和横向扫描，以下的这种方式就是竖向扫描

class Solution {
    public String longestCommonPrefix(String[] strs) {
        StringBuilder commonPrefix = new StringBuilder();
        if (strs == null) {
            return null;
        }
        if (strs.length <= 1) {
            return strs[0];
        }
        int len = strs.length;
        for (int i = 0; i < strs[0].length(); i++) {
            commonPrefix.append(strs[0].charAt(i));
            for (int j = 1; j < len; j++) {
                if (strs[j].length() < commonPrefix.length() || !strs[j].startsWith(commonPrefix.toString())) {
                    commonPrefix.deleteCharAt(i);
                    return commonPrefix.toString();
                }
            }
        }
        return commonPrefix.toString();
    }
}

分治法

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        } else {
            return longestCommonPrefix(strs, 0, strs.length - 1);
        }
    }

    public String longestCommonPrefix(String[] strs, int start, int end) {
        if (start == end) {
            return strs[start];
        } else {
            int mid = (end - start) / 2 + start;
            String lcpLeft = longestCommonPrefix(strs, start, mid);
            String lcpRight = longestCommonPrefix(strs, mid + 1, end);
            return commonPrefix(lcpLeft, lcpRight);
        }
    }

    public String commonPrefix(String lcpLeft, String lcpRight) {
        int minLength = Math.min(lcpLeft.length(), lcpRight.length());       
        for (int i = 0; i < minLength; i++) {
            if (lcpLeft.charAt(i) != lcpRight.charAt(i)) {
                return lcpLeft.substring(0, i);
            }
        }
        return lcpLeft.substring(0, minLength);
    }
}

二分查找

显然，最长公共前缀的长度不会超过字符串数组中的最短字符串的长度。用
minLength 表示字符串数组中的最短字符串的长度，则可以在
[0,minLength] 的范围内通过二分查找得到最长公共前缀的长度。每次取查找范围的中间值
mid，判断每个字符串的长度为
mid 的前缀是否相同，如果相同则最长公共前缀的长度一定大于或等于
mid，如果不相同则最长公共前缀的长度一定小于
mid，通过上述方式将查找范围缩小一半，直到得到最长公共前缀的长度。

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;
    }
}