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题目描述:
- 如果我们可以将小写字母插入模式串 pattern 得到待查询项 query,那么待查询项与给定模式串匹配。(我们可以在任何位置插入每个字符,也可以插入 0 个字符。)
- 给定待查询列表 queries,和模式串 pattern,返回由布尔值组成的答案列表 answer。只有在待查项 queries[i] 与模式串 pattern 匹配时, answer[i] 才为 true,否则为 false。
-  
- 示例 1:
- 输入:queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
- 输出:[true,false,true,true,false]
- 示例:
- "FooBar" 可以这样生成:"F" + "oo" + "B" + "ar"。
- "FootBall" 可以这样生成:"F" + "oot" + "B" + "all".
- "FrameBuffer" 可以这样生成:"F" + "rame" + "B" + "uffer".
- 示例 2:
- 输入:queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
- 输出:[true,false,true,false,false]
- 解释:
- "FooBar" 可以这样生成:"Fo" + "o" + "Ba" + "r".
- "FootBall" 可以这样生成:"Fo" + "ot" + "Ba" + "ll".
- 示例 3:
- 输出:queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
- 输入:[false,true,false,false,false]
- 解释:
- "FooBarTest" 可以这样生成:"Fo" + "o" + "Ba" + "r" + "T" + "est".
-  
- 提示:
- 1 <= queries.length <= 100
- 1 <= queries[i].length <= 100
- 1 <= pattern.length <= 100
- 所有字符串都仅由大写和小写英文字母组成。
- 来源:力扣(LeetCode)
- 链接:https://leetcode-cn.com/problems/camelcase-matching
- 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。
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- class Solution {
- public:
- bool solution(const string& temp1, const string & temp2) {
- int i = 0, j = 0;
- int len1 = temp1.size(), len2 = temp2.size();
- while(i < len1) {
- if(j < len2 && temp1[i] == temp2[j]) {
- i++;
- j++;
- }else if(isupper(temp1[i])) {
- return false;
- }else{
- i++;
- }
- }
- return i == len1 && j == len2;
- }
- vector<bool> camelMatch(vector<string>& queries, string pattern) {
- vector<bool> res;
- for(auto& cha : queries){
- if(solution(cha, pattern))res.push_back(true);
- else{
- res.push_back(false);
- }
- }
- return res;
- }
- };
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参考链接:https://leetcode-cn.com/problems ... -de-zi-fu-chu-nyvs/ |
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