leetcode题解日练--2016.8.26

###不给自己任何借口

今日题目：

1、Ransom Note ； tag:字符串

2、在旋转有序数组中查找最小元素II tag：数组|二分

今日摘录：

草在结它的种子，风在摇它的叶子。我们站着，不说话，就十分美好。

——顾城《门前》

383. Ransom Note | Difficulty: Easy

 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

tag：字符串
题意：判断一个ransom字符串否由magazines中的字符串组成，magazines中的每个字母只能用一次。

思路：用一个vector模拟hash_map就可以了，统计magazines中的字母出现次数，同时减去ransom中每个字母出现的次数，最后判断计数数组中的字母出现的次数有没有负数即可。
1、

class Solution {
public:bool canConstruct(string ransomNote, string magazine) {vector<int>cnt(26,0);for(int i=0;i<ransomNote.size();i++)cnt[ransomNote[i]-'a'+0]--;for(int i=0;i<magazine.size();i++)cnt[magazine[i]-'a'+0]++;for(int i=0;i<26;i++){if(cnt[i]<0)    return false;}return true;}
};

结果：32ms

154. Find Minimum in Rotated Sorted Array II | Difficulty: Hard

Follow up for “Find Minimum in Rotated Sorted Array”:
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?
Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

tag：数组|二分查找
题意：在一个旋转数组中查找最小元素

思路：不断通过和最右元素进行比较，主要思考清楚nums[mid]与nums[right]相等情况下如何处理。

class Solution {
public:int findMin(vector<int>& nums) {int left = 0,right=nums.size()-1;while(left<right){// if(nums[left]<nums[right])  return nums[left];int mid = left+(right-left)/2;if(nums[mid]>nums[right])   left = mid+1;else if (nums[mid] < nums[right])    right = mid;else    right--;}return nums[left];}
};

结果：8ms