给你两个字符串:ransomNote 和 magazine ,判断 ransomNote 能不能由 magazine 里面的字符构成。
如果可以,返回 true ;否则返回 false 。
magazine 中的每个字符只能在 ransomNote 中使用一次。
示例 1:
输入:ransomNote = "a", magazine = "b"
输出:false
示例 2:
输入:ransomNote = "aa", magazine = "ab"
输出:false
示例 3:
输入:ransomNote = "aa", magazine = "aab"
输出:true
提示:
-
1 <= ransomNote.length, magazine.length <= 105 -
ransomNote和magazine由小写英文字母组成pythonclass Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: new_ransomNote = ''.join(sorted(ransomNote)) new_magazine = ''.join(sorted(magazine)) length1 = len(new_ransomNote)//先排序 length2 = len(new_magazine) i,k = 0,0 if(length2<length1)://magazine长度小于ransomNote返回失败 return False else: while(i<length1 and k<length2)://二者数组下标均不可越界 if(new_magazine[k]>new_ransomNote[i]): return False elif(new_magazine[k]==new_ransomNote[i]): i+=1 k+=1 elif(new_magazine[k]<new_ransomNote[i]): k+=1 if(i==length1): return True return False