文章目录
  1. 1.
  2. 2. 题解

Minimum Window Substring

题解

看完Minimum Window Substring【FLAG高频精选面试题讲解】中小姐姐讲的例子,就可以明白此题的思路。

本题为滑动窗口问题,通过hashmap+快慢指针来解决。

针对本题,slow和fast为快慢指针,match_count代表字符串T在字符串S中匹配字符的数目,min_len代表the size of the minimum window in S which will contain all the characters in T,index代表最小滑动窗口的开始地址。

代码是按照Minimum Window Substring【FLAG高频精选面试题讲解】来写的。

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class Solution {
public:
string minWindow(string s, string t) {
unordered_map<char, int> hashmap;
for(int i = 0; i < t.size(); ++i){
if(!hashmap.count(t[i]))
hashmap[t[i]] = 1;
else
hashmap[t[i]]++;
}
int slow = 0, fast = 0, match_count = 0, min_len= INT_MAX, index;
for(fast = 0; fast < s.size(); ++fast){
char ch = s[fast];
if(!hashmap.count(ch))
continue;
int count = hashmap[ch];
hashmap[ch]--;
// match another character
if(count == 1) //1->0
match_count++;
while(match_count == hashmap.size()){
// find a valid substring
if(fast - slow + 1 < min_len){
min_len = fast - slow + 1;
index = slow;
}
char leftmost = s[slow];
slow++;
if(!hashmap.count(leftmost))
continue;
int leftmostCount = hashmap[leftmost];
hashmap[leftmost]++;
if(leftmostCount == 0) // 0->1
match_count--;
}
}

return min_len == INT_MAX ? "" : s.substr(index, min_len);

}
};

参考资料:

  1. Minimum Window Substring【FLAG高频精选面试题讲解】
  2. Here is a 10-line template that can solve most ‘substring’ problems
文章目录
  1. 1.
  2. 2. 题解