你将会得到一份单词表 words,一个字母表 letters (可能会有重复字母),以及每个字母对应的得分情况表 score。
请你帮忙计算玩家在单词拼写游戏中所能获得的「最高得分」:能够由 letters 里的字母拼写出的 任意 属于 words 单词子集中,分数最高的单词集合的得分。
单词拼写游戏的规则概述如下:
玩家需要用字母表 letters 里的字母来拼写单词表 words 中的单词。
可以只使用字母表 letters 中的部分字母,但是每个字母最多被使用一次。
单词表 words 中每个单词只能计分(使用)一次。
根据字母得分情况表score,字母 'a', 'b', 'c', ... , 'z' 对应的得分分别为 score[0], score[1], ..., score[25]。
本场游戏的「得分」是指:玩家所拼写出的单词集合里包含的所有字母的得分之和。
示例 1:
输入:words = ["dog","cat","dad","good"], letters = ["a","a","c","d","d","d","g","o","o"], score = [1,0,9,5,0,0,3,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0]
输出:23
解释:
字母得分为 a=1, c=9, d=5, g=3, o=2
使用给定的字母表 letters,我们可以拼写单词 "dad" (5+1+5)和 "good" (3+2+2+5),得分为 23 。
而单词 "dad" 和 "dog" 只能得到 21 分。
示例 2:
输入:words = ["xxxz","ax","bx","cx"], letters = ["z","a","b","c","x","x","x"], score = [4,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,10]
输出:27
解释:
字母得分为 a=4, b=4, c=4, x=5, z=10
使用给定的字母表 letters,我们可以组成单词 "ax" (4+5), "bx" (4+5) 和 "cx" (4+5) ,总得分为 27 。
单词 "xxxz" 的得分仅为 25 。
示例 3:
输入:words = ["leetcode"], letters = ["l","e","t","c","o","d"], score = [0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0]
输出:0
解释:
字母 "e" 在字母表 letters 中只出现了一次,所以无法组成单词表 words 中的单词。
提示:
1 <= words.length <= 14
1 <= words[i].length <= 15
1 <= letters.length <= 100
letters[i].length == 1
score.length == 26
0 <= score[i] <= 10
words[i] 和 letters[i] 只包含小写的英文字母。
题目是位运算的题,结果没用位运算,用的哈希表和回溯算法。
回溯的思路参考的:
#include <iostream>
#include <unordered_map>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
private:
struct letter {
int score; /* 当前字母值多少分 */
int num; /* 当前字母出现多少次 */
letter() :score(0), num(0) {};
};
public:
int maxScoreWords(vector<string>& words, vector<char>& letters, vector<int>& score) {
GetLettersScore(letters, score);
return backtrack(0, words, charmap);
}
/* 回溯 */
int backtrack(int index, vector<string>& words, unordered_map<char, letter>& charmap) {
if (index >= words.size()) {
return 0;
}
/* 备份,copycharmap:改变过, charmap:未改变过,判断字符串的时候,无非是选和不选 */
unordered_map<char, letter> copycharmap(charmap);
int leftScore, rightScore;
int WordScore = GetCurWordScore(words[index], copycharmap);
/* 选择当前值,累加下一个值*/
if (WordScore == 0) {
leftScore = 0;
}
else {
leftScore = WordScore + backtrack(index + 1, words, copycharmap);
}
/* 不选择当前值 */
rightScore = backtrack(index + 1, words, charmap);
return max(leftScore, rightScore);
}
void GetLettersScore(vector<char>& letters, vector<int>& score) {
for (int i = 0; i < letters.size(); i++) {
int index = letters[i] - 'a'; //score[index] 当前字母的分数
letter cur;
cur.score = score[index];
cur.num = 1;
if (charmap.count(letters[i]) == 1) {
charmap[letters[i]].num++;
}
else {
charmap[letters[i]] = cur;
}
}
/* 显示字母得分 */
//for (int i = 0; i < letters.size(); i++) {
// cout << letters[i] << " : score -> " << charmap[letters[i]].score << " num-> " << charmap[letters[i]].num << endl;
//}
}
private:
/*
哈希表映射:
key: char
value: letter (为什么直接使用letter结构体,而不使用letter*,主要原因是:unorder_map的拷贝构造函数是浅拷贝
*/
unordered_map<char, letter> charmap;
/* 获得当前字符串分数 */
int GetCurWordScore(string& CurWord, unordered_map<char, letter>& charmap) {
int WordScore = 0;
for (char c : CurWord) {
if (charmap.count(c) == 0) {
return 0;
}
else {
//cout << "WordSize:" << WordScore << " Test: char = " << c << " num: " << charmap[c].num << endl;
if (charmap[c].num == 0) {
return 0;
}
else {
WordScore += charmap[c].score;
charmap[c].num--;
}
}
}
//cout << "WordScore:" << WordScore << endl;
return WordScore;
}
};
int main() {
vector<string> words = { "xxxz", "ax", "bx", "cx" };
vector<char> letters = { 'z', 'a', 'b', 'c', 'x', 'x', 'x' };
vector<int> score = { 4, 4, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 10 };
Solution* ps = new Solution();
cout << "final max :" << ps->maxScoreWords(words, letters, score);
return 0;
}