[抄题]:
Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.
Each letter in the magazine string can only be used once in your ransom note.
Note:
You may assume that both strings contain only lowercase letters.
canConstruct("a", "b") -> false canConstruct("aa", "ab") -> false canConstruct("aa", "aab") -> true
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
以为要查找note的起点、终点:没必要,太麻烦,不如遍历,表示每个字母都给查
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
不如遍历,表示每个字母都给查
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
int[26]虽然是数字的数组,但是可以统计字母出现的次数,字符串题常用
[关键模板化代码]:
把字符暂存在数字数组里
int[] chars = new int[26]; //store for (char c : magazine.toCharArray()) { chars[c - 'a']++; }
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public boolean canConstruct(String ransomNote, String magazine) { //cc //ini int[] chars = new int[26]; //store for (char c : magazine.toCharArray()) { chars[c - 'a']++; } //judge for (char c : ransomNote.toCharArray()) { if (--chars[c - 'a'] < 0) { return false; } } //return return true; } }