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LeetCode 392 判断子序列[贪心] HERODING的LeetCode之路
阅读量:386 次
发布时间:2019-03-05

本文共 729 字,大约阅读时间需要 2 分钟。

在这里插入图片描述解题思路:

同时遍历两个字符串,使用双指针,一个遍历t,一个遍历s是否能与t匹配,直到某一个遍历结束,观察s是否遍历结束,结束说明是子序列,否则不是,代码如下:

class Solution {   public:    bool isSubsequence(string s, string t) {           // 定义长度        int len1 = s.length(), len2 = t.length();        if(len1 > len2) {               return false;        }        // 定义索引        int index1 = 0, index2 = 0;        while(index1 < len1 && index2 < len2) {               // 如果当前位置相等,更新s的索引            if(s[index1] == t[index2]) {                   index1 ++;            }            index2 ++;        }        // 返回是否遍历完s        return index1 >= len1;    }};/*作者:heroding链接:https://leetcode-cn.com/problems/is-subsequence/solution/ctan-xin-by-heroding-gbcp/来源:力扣(LeetCode)著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。*/
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