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


    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.
    » Solve this problem

    [解题报告]
    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

    [Code]
    1:    string minWindow(string S, string T) {  
    2: // Start typing your C/C++ solution below
    3: // DO NOT write int main() function
    4: if(S.size() == 0) return "";
    5: if(S.size() < T.size()) return "";
    6: int appearCount[256];
    7: int expectCount[256];
    8: memset(appearCount, 0, 256*sizeof(appearCount[0]));
    9: memset(expectCount, 0, 256*sizeof(appearCount[0]));
    10: for(int i =0; i < T.size(); i++)
    11: {
    12: expectCount[T[i]]++;
    13: }
    14: int minV = INT_MAX, min_start = 0;
    15: int wid_start=0;
    16: int appeared = 0;
    17: for(int wid_end = 0; wid_end< S.size(); wid_end++)
    18: {
    19: if(expectCount[S[wid_end]] > 0)// this char is a part of T
    20: {
    21: appearCount[S[wid_end]]++;
    22: if(appearCount[S[wid_end]] <= expectCount[S[wid_end]])
    23: appeared ++;
    24: }
    25: if(appeared == T.size())
    26: {
    27: while(appearCount[S[wid_start]] > expectCount[S[wid_start]]
    28: || expectCount[S[wid_start]] == 0)
    29: {
    30: appearCount[S[wid_start]]--;
    31: wid_start ++;
    32: }
    33: if(minV > (wid_end - wid_start +1))
    34: {
    35: minV = wid_end - wid_start +1;
    36: min_start = wid_start;
    37: }
    38: }
    39: }
    40: if(minV == INT_MAX) return "";
    41: return S.substr(min_start, minV);
    42: }

    [已犯错误]
    1. Line 8&9
    不熟悉这个api,最初写成了

    memset(expectCount, 0, 256);

    结果老是出问题,检查了很多遍logic,也没发现有问题,最后还是放到VS下debug才发现原来是地址空间大小没有传对。正确的应该是:

    memset(expectCount, 0, 256*sizeof(appearCount[0]));






  • 相关阅读:
    ApacheTika解析Word文档
    Oracle中生成随机数的函数
    实现Android和PC之间的蓝牙通信(转载)
    以C#编写的Socket服务器的Android手机聊天室Demo
    初识SVM
    一些编程的小练手
    c#连接MYSQL数据库的两种方法
    Doxygen+Graphviz用来画程序结构图
    C#+MYSQL数据库操作(附源码)
    [转]Use PowerShell to Manage Lists, Views, and Items in SharePoint(使用PowerShell管理列表、视图、列表项)
  • 原文地址:https://www.cnblogs.com/codingtmd/p/5078985.html
Copyright © 2011-2022 走看看