【LeetCode】每日一题（8.6）

回文对

给定一组唯一的单词， 找出所有不同 的索引对(i, j)，使得列表中的两个单词， words[i] + words[j] ，可拼接成回文串。

示例 1:

输入: ["abcd","dcba","lls","s","sssll"]
输出: [[0,1],[1,0],[3,2],[2,4]] 
解释: 可拼接成的回文串为 ["dcbaabcd","abcddcba","slls","llssssll"]

示例 2:

输入: ["bat","tab","cat"]
输出: [[0,1],[1,0]] 
解释: 可拼接成的回文串为 ["battab","tabbat"]

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/palindrome-pairs
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路

枚举

代码实现

class Solution {
public:
    struct node {
        int ch[26];
        int flag;
        node() {
            flag = -1;
            memset(ch, 0, sizeof(ch));
        }
    };

    vector<node> tree;

    void insert(string& s, int id) {
        int len = s.length(), add = 0;
        for (int i = 0; i < len; i++) {
            int x = s[i] - 'a';
            if (!tree[add].ch[x]) {
                tree.emplace_back();
                tree[add].ch[x] = tree.size() - 1;
            }
            add = tree[add].ch[x];
        }
        tree[add].flag = id;
    }

    int findWord(string& s, int left, int right) {
        int add = 0;
        for (int i = right; i >= left; i--) {
            int x = s[i] - 'a';
            if (!tree[add].ch[x]) {
                return -1;
            }
            add = tree[add].ch[x];
        }
        return tree[add].flag;
    }

    bool isPalindrome(string& s, int left, int right) {
        int len = right - left + 1;
        for (int i = 0; i < len / 2; i++) {
            if (s[left + i] != s[right - i]) {
                return false;
            }
        }
        return true;
    }

    vector<vector<int>> palindromePairs(vector<string>& words) {
        tree.emplace_back(node());
        int n = words.size();
        for (int i = 0; i < n; i++) {
            insert(words[i], i);
        }
        vector<vector<int>> ret;
        for (int i = 0; i < n; i++) {
            int m = words[i].size();
            for (int j = 0; j <= m; j++) {
                if (isPalindrome(words[i], j, m - 1)) {
                    int left_id = findWord(words[i], 0, j - 1);
                    if (left_id != -1 && left_id != i) {
                        ret.push_back({i, left_id});
                    }
                }
                if (j && isPalindrome(words[i], 0, j - 1)) {
                    int right_id = findWord(words[i], j, m - 1);
                    if (right_id != -1 && right_id != i) {
                        ret.push_back({right_id, i});
                    }
                }
            }
        }
        return ret;
    }
};

前两天都没写每日一题，接下来两天也不一定写，这题也是抄的，还是专心把“回溯算法”给办了。