给定一个单词列表 words 和一个整数 k ,返回前 k 个出现次数最多的单词。
返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率, 按字典顺序 排序。
示例 1:
输入: words = ["i", "love", "leetcode", "i", "love", "coding"], k = 2
输出: ["i", "love"]
解析: "i" 和 "love" 为出现次数最多的两个单词,均为2次。
注意,按字母顺序 "i" 在 "love" 之前。
示例 2:
输入: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
输出: ["the", "is", "sunny", "day"]
解析: "the", "is", "sunny" 和 "day" 是出现次数最多的四个单词,
出现次数依次为 4, 3, 2 和 1 次。
注意:
1 <= words.length <= 5001 <= words[i] <= 10words[i] 由小写英文字母组成。k 的取值范围是 [1, 不同 words[i] 的数量]思路:
首先想到了map做键值对映射,记录单词的出现次数,最后造个数组排序就行,但是面试一紧张忘了map怎么用了,只能另辟蹊径,改用C++结构体,分别记录单词,出现次数,是否是第一次出现
结构体解法:
- struct ac {
- string word;
- int num;
- int flag;
- } a[505];
-
-
- bool cmp(ac a, ac b) {
- if (a.num == b.num) {
- return a.word < b.word;
- }
- return a.num > b.num;
- }
- class Solution {
- public:
-
- vector
topKFrequent(vector& words, int k) { - vector
res; - int i, j;
- for (i = 0; i < words.size(); i++) {
- // cout<
- a[i].word = words[i];
- a[i].num = 0;
- a[i].flag = 0;
- }
-
- for (i = 0; i < words.size(); i++) {
- for (j = 0; j < words.size(); j++) {
- if (a[j].word == words[i]) {
-
- if (a[j].num != 0)
- a[j].flag = 1;
- a[j].num++;
-
- }
- }
- }
-
- sort(a, a + words.size(), cmp);
-
- for (i = 0; i < words.size(); i++) {
-
- cout << "--" << a[i].word << " " << a[i].num << endl;
- }
-
- for (i = 0; i < words.size(); i++) {
- if (i != 0 && a[i].word == a[i - 1].word)
- continue;
- res.push_back(a[i].word);
- // cout << a[i].word << endl;
- k--;
- if (k <= 0)
- break;
- }
-
- return res;
- }
- };
map解法:
- struct ac {
- string word;
- int num;
- int flag;
- } a[505];
-
-
- bool cmp(ac a, ac b) {
- if (a.num == b.num) {
- return a.word < b.word;
- }
- return a.num > b.num;
- }
- class Solution {
- public:
-
- vector
topKFrequent(vector& words, int k) { - vector
res; - int i, j;
-
- map
int> mp; -
- // for (i = 0; i < 6; i++) {
- // string s;
- // cin >> s;
- // words.push_back(s);
- // }
-
- for (i = 0; i < words.size(); i++) {
- // cout<
- mp[words[i]]++;
- }
-
- i = 0;
- map
int>::iterator it; - for (it = mp.begin(); it != mp.end(); it++) {
- a[i].word = it->first;
- a[i].num = it->second;
- i++;
- }
-
- sort(a, a + i, cmp);
-
- // for (i = 0; i < words.size(); i++) {
- // cout << "--" << a[i].word << " " << a[i].num << endl;
- // }
-
- for (i = 0; i < k; i++) {
- res.push_back(a[i].word);
- }
-
- return res;
- }
- };