力扣解法汇总1023. 驼峰式匹配

目录链接:

力扣编程题-解法汇总_分享+记录-CSDN博客

GitHub同步刷题项目:

https://github.com/September26/java-algorithms

原题链接:力扣


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. 1 <= queries.length <= 100
  2. 1 <= queries[i].length <= 100
  3. 1 <= pattern.length <= 100
  4. 所有字符串都仅由大写和小写英文字母组成。

解题思路:

* 解题思路:
* isMatch来判断对应的字符是否满足,用两个int值index1和index2记录位置。
* 如果chars1[index1]==chars2[index2],则index1++,index2。
* 否则index1++。等于说只挪动如果chars1中的位置,如果这个位置是大些字母,则直接返回false。
* 遍历完chars1后,如果发现还有未遍历的chars2,则说明没有覆盖全字符串pattern,则false。

代码:

public class Solution1023 {

    public List camelMatch(String[] queries, String pattern) {
        List list = new ArrayList<>();
        Arrays.stream(queries).forEach(s -> list.add(isMatch(s, pattern)));
        return list;
    }

    public Boolean isMatch(String querie, String pattern) {
        char[] chars1 = querie.toCharArray();
        char[] chars2 = pattern.toCharArray();
        int index1 = 0;
        int index2 = 0;
        while (index1 < chars1.length && index2 < chars2.length) {
            char queriChar = chars1[index1];
            if (queriChar == chars2[index2]) {
                index1++;
                index2++;
                continue;
            }
            index1++;
            if (queriChar >= 'A' && queriChar <= 'Z') {
                return false;
            }
        }
        if (index2 < chars2.length) {
            return false;
        }
        while (index1 < chars1.length) {
            char queriChar = chars1[index1];
            if (queriChar >= 'A' && queriChar <= 'Z') {
                return false;
            }
            index1++;
        }
        return true;
    }
}

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