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LeetCode题解（Java）：14-最长公共前缀

2020年07月13日  | 移动技术网IT编程  | 我要评论

14. 最长公共前缀

https://leetcode-cn.com/problems/longest-common-prefix/

解法一：横向比较

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs.length == 0) return "";
        String prefix = strs[0];
        for (int i = 1; i < strs.length; i++) {
        	int k = 0;
            String candidate = strs[i];
            int minLength = Math.min(prefix.length(), candidate.length());
            while (k < minLength && prefix.charAt(k) == candidate.charAt(k)) {
                k++;
            }
            prefix = prefix.substring(0, k);
        }
        return prefix;
    }
}

解法二：纵向比较

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs.length == 0) return "";
        String prefix = strs[0];
        int i = 0;
        while (i < prefix.length()) {
            char ch = prefix.charAt(i);
            for (int j = 1; j < strs.length; j++) {
                String candidate = strs[j];
                if (i > candidate.length() - 1 || ch != candidate.charAt(i)) {
                    return prefix.substring(0, i);
                }
            }
            i++;
        }
        return prefix;
    }
}

解法三：分治

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

    public String dac(String[] strs, int lo, int hi) {
        if (lo == hi) return strs[lo];
        int mid = lo + ((hi - lo) >> 1);
        String left = dac(strs, lo, mid);
        String right = dac(strs, mid + 1, hi);
        return lcp(left, right);
    }

    public String lcp(String s1, String s2) {
        int minL = Math.min(s1.length(), s2.length());
        for (int i = 0; i < minL; i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                return s1.substring(0, i);
            }
        }
        return s1.substring(0, minL);
    }
}

解法四：二分查找

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs.length == 0) return "";
        int minLen = Integer.MAX_VALUE;
        for (String s : strs) {
            minLen = Math.min(minLen, s.length());
        }
        int lo = 0, hi = minLen - 1;
        while (lo <= hi) {
            int mid = lo + ((hi - lo) >> 1);
            if (check(strs, lo, mid)) {// [lo, mid] is same
                lo = mid + 1;
            } else {
                hi = mid - 1;
            }
        }
        return strs[0].substring(0, lo);
    }

    public boolean check(String[] strs, int lo, int hi) {
        String prefix = strs[0];
        while (lo <= hi) {
            for (int i = 1; i < strs.length; i++) {
                if (prefix.charAt(lo) != strs[i].charAt(lo)) {
                    return false;
                }
            }
            lo++;
        }
        return true;
    }
}

解法五：indexOf() || startWith()

class Solution {
    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].startsWith(prefix)) {
            //     prefix = prefix.substring(0, prefix.length() - 1);
            // }
            while (0 != strs[i].indexOf(prefix)) {
                prefix = prefix.substring(0, prefix.length() - 1);
            }
        }
        return prefix;
    }
}

本文地址：https://blog.csdn.net/qq_29511515/article/details/107289668

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