leetcode 383. Ransom Note
2016-09-11 11:06
477 查看
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.
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
class Solution { public: bool canConstruct(string ransomNote, string magazine) { map<char,int> mpr,mpm; map<char,int>::iterator itr,itm; int i ,j; for(i=0; i < ransomNote.length();++i) { mpr[ransomNote[i]]++; } for(j = 0; j<magazine.length();++j) { mpm[magazine[j]]++; } if(mpr.size() > mpm.size()) return false; itr = mpr.begin();itm = mpm.begin(); while(itr != mpr.end() && itm != mpm.end()) { char tmp = itr->first; map<char,int>::iterator t; if(mpm.find(tmp) == mpm.end()) return false; if((int)(itr->second) > (int)(mpm.find(tmp)->second)) return false; else itr++; } return true; } };
相关文章推荐
- leetcode(383. Ransom Note)
- Leetcode 383. Ransom Note 构造字符串 解题报告
- leetcode_383. Ransom Note-近似子串问题
- leetcode(85).383. Ransom Note
- LeetCode之383. Ransom Note
- LeetCode 383. Ransom Note
- LeetCode 383. Ransom Note 解题报告
- Leetcode——383. Ransom Note
- leetCode 383. Ransom Note 字符串
- LeetCode 383. Ransom Note
- LeetCode-383. Ransom Note
- leetcode题解-58. Length of Last Word && 67. Add Binary && 383. Ransom Note
- leetcode 383. Ransom Note 勒索信
- 【Leetcode】383. Ransom Note
- LeetCode笔记:383. Ransom Note
- Leetcode383. Ransom Note
- LeetCode - 383. Ransom Note
- LeetCode 383. Ransom Note(java)
- LeetCode-383. Ransom Note
- leetcode 383. Ransom Note