Leetcode 76. Minimum Window Substring (python+CPP)

题目

解法:sliding window

直接上leetcode 官方solution

• We start with two pointers, leftleft and rightright initially pointing to the first element of the string SS.
• We use the rightright pointer to expand the window until we get a desirable window i.e. a window that contains all of the characters of TT.
• Once we have a window with all the characters, we can move the left pointer ahead one by one. If the window is still a desirable one we keep on updating the minimum window size.
• If the window is not desirable any more, we repeat step ; 2step2 onwards.
  python代码如下：

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        if not t or not s:
            return ''
        
        dict_t = Counter(t)
        required = len(dict_t)
        l,r = 0,0
        formed = 0
        window_counts = {}
        ans = float('inf'),None,None
        
        while r<len(s):
            character = s[r]
            window_counts[character] = window_counts.get(character,0)+1
            
            if character in dict_t and window_counts[character] == dict_t[character]:
                formed += 1
            
            while l<=r and formed==required:
                character = s[l]
                
                if r-l+1 < ans[0]:
                    ans = (r-l+1,l,r)
                    
                window_counts[character] -= 1
                if character in dict_t and window_counts[character]<dict_t[character]:
                    formed -= 1
                
                l += 1
            
            r += 1
        
        return '' if ans[0]==float('inf') else s[ans[1]:ans[2]+1]

C++代码如下：

class Solution {
public:
    string minWindow(string s, string t) {
        vector<int> count_t(128,0);
        vector<int> count_window(128,0);
        int required = 0;
        
        for(int i=0;i<t.size();++i){
            if(count_t[t[i]]==0) ++required;
            ++count_t[t[i]];
        }
        
        int l=0, r=0,formed=0;
        int window_size = s.size()+1,final_l=0,final_r=0;
        
        while (r<s.size()){
            count_window[s[r]] = count_window[s[r]]+1;
            if (count_window[s[r]]==count_t[s[r]]) ++formed;
            
            while (l<=r && formed==required){
                if (r-l+1 < window_size) {
                    window_size = r-l+1;
                    final_l = l;
                    final_r = r;
                }
                
                --count_window[s[l]];
                if (count_window[s[l]] < count_t[s[l]]) --formed;
                
                ++l;
            }
            ++r;
        }
        return window_size == s.size()+1? "":s.substr(final_l,final_r-final_l+1);
        
        
    }
};

参考链接