PAT(Advanced) 1092 To Buy or Not to Buy C++实现_to buy or not to buyc++-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_44486439/article/details/113400012

本文介绍了一道算法题目，如何利用C++ unordered_map数据结构解决字符串包含问题，计算给定字符串中多余的字符和目标字符串缺少的字符数量。通过哈希表快速比较字符计数，适用于字符计数不需保持有序的情况。

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PAT(Advanced) 1092 To Buy or Not to Buy C++实现

题目链接

1092 To Buy or Not to Buy

题目大意

给定字符串和目标字符串，若给定字符串中包含目标字符串中所有的字符，则输出Yes并输出给定字符串中多余的字符的个数，否则输出No并输出除去给定字符串包含的字符后目标字符串中仍缺少的字符串的个数。

算法思路

借助C++ STL容器unordered_map，将给定字符串和目标字符串中所有的字符计数，遍历目标字符串中的字符计数值，与给定字符串的字符计数值做差运算，最后若目标字符串中已经没有字符则说明满足条件输出给定字符串中多余的字符个数，否则输出目标字符串中缺少的个数。本题map容器一样可以实现，map容器基于红黑树，迭代器遍历结果为有序序列，而unordered_map容器基于哈希表，迭代器遍历结果为无序序列，本题不要求迭代结果的有序，使用unorder_map会更快，unorder_map要求编译器版本高于C++ 11标准。

AC代码

/*
author : eclipse
email  : eclipsecs@qq.com
time   : Fri Jan 29 16:46:04 2021
*/
#include <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[]) {
    string givenString;
    string targetString;
    cin >> givenString >> targetString;
    unordered_map<char, int> givenBeadsCount;
    unordered_map<char, int> targetBeadsCount;

    for (int i = 0; i < givenString.size(); i++) {
        if (givenBeadsCount.count(givenString[i])) {
            givenBeadsCount[givenString[i]]++;            
        } else {
            givenBeadsCount[givenString[i]] = 1;
        }
    }

    for (int i = 0; i < targetString.size(); i++) {
        if (targetBeadsCount.count(targetString[i])) {
            targetBeadsCount[targetString[i]]++;
        } else {
            targetBeadsCount[targetString[i]] = 1;
        }
    }

    int givenStringRest = givenString.size();
    int targetStringRest = targetString.size();

    for (unordered_map<char, int>::iterator it = targetBeadsCount.begin(); it != targetBeadsCount.end(); it++) {
        if (givenBeadsCount.count(it->first)) {            
            if (givenBeadsCount[it->first] >= it->second) {
                givenStringRest -= it->second;
                targetStringRest -= it->second;
            } else {
                givenStringRest -= givenBeadsCount[it->first];
                targetStringRest -= givenBeadsCount[it->first];
            }
        }
    }

    printf("%s %d", targetStringRest == 0 ? "Yes" : "No", targetStringRest == 0 ? givenStringRest : targetStringRest);

    return 0;
}