zoukankan      html  css  js  c++  java
  • [LeetCode] Minimum Window Substring

    Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

    For example,
    S = "ADOBECODEBANC"
    T = "ABC"

    Minimum window is "BANC".

    Note:
    If there is no such window in S that covers all characters in T, return the emtpy string "".

    If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

    Hide Tags
     Hash Table Two Pointers String
     

    思路:

    双指针,动态维护一个区间。尾指针不断往后扫,当扫到有一个窗口包含了所有 T 的字符后,
    然后再收缩头指针,直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

    时间复杂度 O(n),空间复杂度 O(1)

    class Solution {
        public:
            string minWindow(string S, string T)
            {
                if(S.empty() || T.empty() || S.size() < T.size())
                    return string();
    
                vector<int> expect(256, 0);
                vector<int> appear(256, 0);
    
                for(int i = 0; i < T.size(); i++)
                {   
                    expect[T[i]] ++; 
                }   
    
                int minWidth = INT_MAX, min_start = 0; //
                int win_start = 0;
                int appearCharCnt = 0;
                for(int win_end = 0; win_end < S.size(); win_end++)
                {   
                    if(expect[S[win_end]] > 0) // this char is part of T
                    {   
                        appear[S[win_end]]++;
                        if(appear[S[win_end]] <= expect[S[win_end]])
                            appearCharCnt ++; 
                    }   
                    //cout << "appearCharCnt	" <<appearCharCnt<< endl;
                    if(appearCharCnt == T.size())
                    {   
                        // shrink the start
                        while (appear[S[win_start]] > expect[S[win_start]]
                                || expect[S[win_start]] == 0) {
                            appear[S[win_start]]--;
                            win_start++;
                        }
                        if ((win_end - win_start + 1) < minWidth) {
                            minWidth = win_end - win_start + 1;
                            min_start = win_start;
                            //cout << "min_start	" <<min_start << endl;
                            //cout << "min_width	" <<minWidth<< endl;
                        }
                    }
                }
    
                if (minWidth == INT_MAX)
                    return "";
                else
                    return S.substr(min_start, minWidth);
    
    
            }
    };

    精简一下条件判断

    class Solution {
        public:
            string minWindow(string S, string T)
            {
                if(S.empty() || T.empty() || S.size() < T.size())
                    return string();
    
                vector<int> expect(256, 0);
                vector<int> appear(256, 0);
    
                for(int i = 0; i < T.size(); i++)
                {
                    expect[T[i]] ++;
                }
    
                int minWidth = INT_MAX, min_start = 0; 
                int win_start = 0;
                int appearCharCnt = 0;
                for(int win_end = 0; win_end < S.size(); win_end++)
                {   
                    appear[S[win_end]]++;
                    if(appear[S[win_end]] <= expect[S[win_end]])
                        appearCharCnt ++; 
                    //cout << "appearCharCnt	" <<appearCharCnt<< endl;
                    if(appearCharCnt == T.size())
                    {   
                        // shrink the win start
                        while (appear[S[win_start]] > expect[S[win_start]]
                              ) { 
                            appear[S[win_start]]--;
                            win_start++;
                        }   
                        if ((win_end - win_start + 1) < minWidth) {
                            minWidth = win_end - win_start + 1;
                            min_start = win_start;
                            //cout << "min_start	" <<min_start << endl;
                            //cout << "min_width	" <<minWidth<< endl;
                        }
                    }
                }
    
                if (minWidth == INT_MAX)
                    return "";
                else
                    return S.substr(min_start, minWidth);
    
    
            }
    };
  • 相关阅读:
    大话设计模式之代理模式
    大话设计模式之装饰者模式
    策略模式与简单工厂模式
    一个简单的使用Quartz和Oozie调度作业给大数据计算平台执行
    oozie JAVA Client 编程提交作业
    HashMap分析及散列的冲突处理
    cmp排序hdoj 1106排序
    定义member【C++】cstddef中4个定义
    目录启动CXF启动报告LinkageError异常以及Java的endorsed机制
    算法代码[置顶] 机器学习实战之KNN算法详解
  • 原文地址:https://www.cnblogs.com/diegodu/p/4329944.html
Copyright © 2011-2022 走看看