[Leetcode] 524. Longest Word in Dictionary through Deleting 解题报告_python 524. longest word in dictionary through del-优快云博客

本文介绍了一种算法，用于从给定字符串中找出字典中最长的子序列字符串。通过先对字典中的字符串按长度及字典序排序，再逐一检查是否为原字符串的子序列来解决问题。

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题目：

Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

Example 1:

Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]

Output: 
"apple"

Example 2:

Input:
s = "abpcplea", d = ["a","b","c"]

Output: 
"a"

Note:

All the strings in the input will only contain lower-case letters.
The size of the dictionary won't exceed 1,000.
The length of all the strings in the input won't exceed 1,000.

思路：

还是判断一个字符串是否是另一个字符串的子序列的问题：我们首先将字典中的字符串进行排序，使得长度长的位于前面，如果长度相同，则按照字符串的字典序排列。然后我们遍历排序后的字典，一旦发现字典中的某个字符串是s的子串，就返回字典中的这个字符串。如果都没有找到，就返回空串。算法的时间复杂度是O(nlogn)，空间复杂度是O(1)。

代码：

class Solution {
public:
    string findLongestWord(string s, vector<string>& d) {
        sort(d.begin(), d.end(), StringComp);
        for (int i = 0; i < d.size(); ++i) {
            if (isSubString(d[i], s)) {
                return d[i];
            }
        }
        return "";
    }
private:
    struct StringCompare {
        bool operator() (const string &s1, const string &s2) const {
            if (s1.length() != s2.length()) {
                return s1.length() > s2.length();
            }
            else {
                return s1 < s2;
            }
        }
    } StringComp;
    bool isSubString(string &s1, string &s2) {  // determine whether s1 is the substring of s2
        int i = 0, j = 0;
        for (; i < s1.length(); ++i) {
            while (j < s2.length() && s1[i] != s2[j]) {
                ++j;
            }
            if (j == s2.length()) {
                break;
            }
            else {
                ++j;
            }
        }
        return i == s1.length();
    }
};