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.

    class Solution {
    public:
        int sub(char c){
            if(c>='a'&&c<='z'){
                return c-'a';
            }
            else if(c>='A'&&c<='Z')
            {
                return c-'A'+26;
            }
            else return -1;
        }
        string minWindow(string S, string T) {
            // Note: The Solution object is instantiated only once and is reused by each test case.
            vector<int> count,count2;
            count.resize(52,0);
            count2.resize(52,0);
            for(int i=0;i<T.length();i++){
                int ind=sub(T[i]);
                if(ind!=-1)count[ind]++;
            }
            int dis=0;
            for(int i=0;i<52;i++){
                dis+=count[i];
            }
            int mins=-1,mine;
            int start=0,end=0;
            int ind=sub(S[end]);
            if(ind!=-1){
                count2[ind]++;
                if(count2[ind]<=count[ind])dis--;
            }
            while(true){
                if(dis!=0){
                    end++;
                    if(end==S.length())break;
                    int ind=sub(S[end]);
                    if(ind!=-1){
                        count2[ind]++;
                        if(count2[ind]<=count[ind])dis--;
                    }
                }
                else{
                    if(mins==-1||end-start<mine-mins){
                        mins=start;
                        mine=end;
                    }
                    int ind=sub(S[start]);
                    if(ind!=-1){
                        count2[ind]--;
                        if(count2[ind]<count[ind]){
                            dis++;
                        }
                    }
                    start++;
                    if(start>end)break;
                }
            }
            if(mins==-1)return "";
            return S.substr(mins,mine-mins+1);
        }
    };
    View Code
  • 相关阅读:
    线程(java课堂笔记)
    java中的各种流(老师的有道云笔记)
    面向对象(java菜鸟的课堂笔记)
    泛型(java菜鸟的课堂笔记)
    我做的第一个程序(菜鸟的java课堂笔记)
    java中的一些规则(菜鸟的课堂笔记)
    一位菜鸟的java 最基础笔记
    spatial index (空间索引)
    hadoop 的疑问
    numpy 矩阵的运算
  • 原文地址:https://www.cnblogs.com/superzrx/p/3354599.html
Copyright © 2011-2022 走看看