标签：turn 因此 break return input ranking while pos 利用
问题：
给出多次，字母排名次的名次列表。
求的综合名次的结果。（如果两字母分值相同，则按照字母序排列）
Example 1: Input: votes = ["ABC","ACB","ABC","ACB","ACB"] Output: "ACB" Explanation: Team A was ranked first place by 5 voters. No other team was voted as first place so team A is the first team. Team B was ranked second by 2 voters and was ranked third by 3 voters. Team C was ranked second by 3 voters and was ranked third by 2 voters. As most of the voters ranked C second, team C is the second team and team B is the third. Example 2: Input: votes = ["WXYZ","XYZW"] Output: "XWYZ" Explanation: X is the winner due to tie-breaking rule. X has same votes as W for the first position but X has one vote as second position while W doesn‘t have any votes as second position. Example 3: Input: votes = ["ZMNAGUEDSJYLBOPHRQICWFXTVK"] Output: "ZMNAGUEDSJYLBOPHRQICWFXTVK" Explanation: Only one voter so his votes are used for the ranking. Example 4: Input: votes = ["BCA","CAB","CBA","ABC","ACB","BAC"] Output: "ABC" Explanation: Team A was ranked first by 2 voters, second by 2 voters and third by 2 voters. Team B was ranked first by 2 voters, second by 2 voters and third by 2 voters. Team C was ranked first by 2 voters, second by 2 voters and third by 2 voters. There is a tie and we rank teams ascending by their IDs. Example 5: Input: votes = ["M","M","M","M"] Output: "M" Explanation: Only team M in the competition so it has the first rank. Constraints: 1 <= votes.length <= 1000 1 <= votes[i].length <= 26 votes[i].length == votes[j].length for 0 <= i, j < votes.length. votes[i][j] is an English upper-case letter. All characters of votes[i] are unique. All the characters that occur in votes[0] also occur in votes[j] where 1 <= j < votes.length.
解法：
名次得分法：
第一位得分最高，第二位得分次之，以此类推，最后一位得分最小为1
累计各个字母的累计分值，
按照分值从大到小，最大分值的字母则综合排名为第一位。
vector<vector<int>> count(26, vector<int>(27));
我们利用上述结构。记录A～Z字母出现的次数
count[‘A‘][0~25]表示字母A出现在0～25位上的得分。
最后排序需要另外记录该字母，且若两字母分值相同，按照字母顺序计算分值
因此多加一位count[‘A‘][26]，保存该字母。
（这样，即保存了该字母是哪个字母，又将字母序也放入分值计算中）
我们遍历所有排序中所有字母的出现，同时计算分值，计入count中。
?? 注意，由于最终按照出现次数最多，排序在前面，
我们若想使用默认从小到大的sort排序，那么，我们使出现次数最多，得分值最小，这样去计算分值。
即，出现一次，值--，那么出现越多，值越小。
最后，对count进行默认sort排序。
从头到尾依次输出到res中。
代码参考：
1 class Solution { 2 public: 3 string rankTeams(vector<string>& votes) { 4 string res; 5 int N=votes.size(); 6 int POINT=votes[0].size(); 7 if(N==1 || POINT==1) return votes[0]; 8 vector<vector<int>> count(26, vector<int>(27)); 9 for(char c:votes[0]){ 10 count[c-‘A‘][26]=c; 11 } 12 for(string s:votes){ 13 int po=POINT; 14 for(int i=0; i<POINT; i++){ 15 count[s[i]-‘A‘][i]--; 16 } 17 } 18 sort(count.begin(), count.end()); 19 for(int i=0; i<POINT; i++){ 20 res+=count[i][26]; 21 } 22 return res; 23 } 24 };
标签：turn 因此 break return input ranking while pos 利用
原文地址：https://www.cnblogs.com/habibah-chang/p/13284175.html
