class Solution {
public:
vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
}
};
6347. 统计范围内的元音字符串数
给你一个下标从 0 开始的字符串数组 words 以及一个二维整数数组 queries 。
每个查询 queries[i] = [li, ri] 会要求我们统计在 words 中下标在 li 到 ri 范围内（包含 这两个值）并且以元音开头和结尾的字符串的数目。
返回一个整数数组，其中数组的第 i 个元素对应第 i 个查询的答案。
注意：元音字母是 'a'、'e'、'i'、'o' 和 'u' 。
示例 1：
输入：words = ["aba","bcb","ece","aa","e"], queries = [[0,2],[1,4],[1,1]] 输出：[2,3,0] 解释：以元音开头和结尾的字符串是 "aba"、"ece"、"aa" 和 "e" 。 查询 [0,2] 结果为 2（字符串 "aba" 和 "ece"）。 查询 [1,4] 结果为 3（字符串 "ece"、"aa"、"e"）。 查询 [1,1] 结果为 0 。 返回结果 [2,3,0] 。
示例 2：
输入：words = ["a","e","i"], queries = [[0,2],[0,1],[2,2]] 输出：[3,2,1] 解释：每个字符串都满足这一条件，所以返回 [3,2,1] 。
提示：
1 <= words.length <= 1051 <= words[i].length <= 40words[i] 仅由小写英文字母组成sum(words[i].length) <= 3 * 1051 <= queries.length <= 1050 <= queries[j][0] <= queries[j][1] < words.length原站题解
javascript 解法, 执行用时: 140 ms, 内存消耗: 67.8 MB, 提交时间: 2023-11-07 07:23:19
/**
* @param {string[]} words
* @param {number[][]} queries
* @return {number[]}
*/
var vowelStrings = function(words, queries) {
let n = words.length;
let prefixSums = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
let value = isVowelString(words[i]) ? 1 : 0;
prefixSums[i + 1] = prefixSums[i] + value;
}
let ans = [];
for (let i = 0; i < queries.length; i++) {
let start = queries[i][0], end = queries[i][1];
ans.push(prefixSums[end + 1] - prefixSums[start]);
}
return ans;
}
function isVowelString(word) {
return isVowelLetter(word[0]) && isVowelLetter(word[word.length - 1]);
}
function isVowelLetter(c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
java 解法, 执行用时: 3 ms, 内存消耗: 81.9 MB, 提交时间: 2023-11-07 07:23:00
class Solution {
public int[] vowelStrings(String[] words, int[][] queries) {
int n = words.length;
int[] prefixSums = new int[n + 1];
for (int i = 0; i < n; i++) {
int value = isVowelString(words[i]) ? 1 : 0;
prefixSums[i + 1] = prefixSums[i] + value;
}
int q = queries.length;
int[] ans = new int[q];
for (int i = 0; i < q; i++) {
int start = queries[i][0], end = queries[i][1];
ans[i] = prefixSums[end + 1] - prefixSums[start];
}
return ans;
}
public boolean isVowelString(String word) {
return isVowelLetter(word.charAt(0)) && isVowelLetter(word.charAt(word.length() - 1));
}
public boolean isVowelLetter(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
}
cpp 解法, 执行用时: 132 ms, 内存消耗: 62.3 MB, 提交时间: 2023-02-06 10:11:44
class Solution {
public:
vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
int n = queries.size();
vector<int> ans(n);
int m = words.size();
vector<int> flag(m);
for (int i = 0; i < m; ++i) {
string str = words[i];
int len = str.size();
if ((str[0] == 'a' || str[0] == 'e' || str[0] == 'i' || str[0] == 'o' || str[0] == 'u') && (str[len-1] == 'a' || str[len-1] == 'e' || str[len-1] == 'i' || str[len-1] == 'o' || str[len-1] == 'u'))
flag[i] = 1;
else flag[i] = 0;
}
vector<int> pre(m+1);
for (int i = 1; i <= m; ++i) {
pre[i] = pre[i-1] + flag[i-1];
}
for (int i = 0; i < n; ++i) {
int num = 0;
ans[i] = pre[queries[i][1] + 1] - pre[queries[i][0]];
}
return ans;
}
};
golang 解法, 执行用时: 112 ms, 内存消耗: 19.1 MB, 提交时间: 2023-02-06 10:10:26
func vowelStrings(words []string, queries [][]int) []int {
sum := make([]int, len(words)+1)
for i, w := range words {
sum[i+1] = sum[i]
if strings.Contains("aeiou", w[:1]) && strings.Contains("aeiou", w[len(w)-1:]) {
sum[i+1]++
}
}
ans := make([]int, len(queries))
for i, q := range queries {
ans[i] = sum[q[1]+1] - sum[q[0]]
}
return ans
}
python3 解法, 执行用时: 80 ms, 内存消耗: 45.9 MB, 提交时间: 2023-02-06 10:09:53
'''
前缀和
'''
class Solution:
def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
s = list(accumulate((w[0] in "aeiou" and w[-1] in "aeiou" for w in words), initial=0))
return [s[r + 1] - s[l] for l, r in queries]