[CareerCup][Google Interview] 排序字符串

Sort an input string according to the given order string. There might be characters in input string that are not present in the order string and vice-versa. The characters of input string that are not present in the order string should come at the end of the output string in any order. Write code...

Example: Sort("abdcdfs","dacg");

output: ddacfbs

方法1：考虑用一个图来保存字符的大小关系，这样排序字符串的时候可以根据各各字符的大小关系来排序。构建图O(m^2)，m为关系字符串长度。排序O(nlogn)

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

bool isSmaller[256][256];

bool comp(const char &lhs, const char &rhs)
{
    return isSmaller[lhs][rhs];
}

int main()
{
    memset(isSmaller, false, sizeof(isSmaller));

    char s[] = "abdcdfs";
    string order = "dacg";

    for(int i = 0; i < order.size(); i++)
    {
        for(int j = i + 1; j < order.size(); j++)
            isSmaller[order[i]][order[j]] = true;
        for(int k = 0; k < 256; k++)
            if (!isSmaller[k][order[i]])
                isSmaller[order[i]][k] = true;
    }

    int len = strlen(s);
    
    sort(s, s + len, comp);

    cout << s << endl;
}

方法2：可以用一个hash表来保存在order中的字符出现个数，然后把未在order中出现的字符放后面。整个算法复杂度O(n)

#include <iostream>
#include <map>
#include <string>
using namespace std;

void Sort(string &s, string &order)
{
    int count[256];

    memset(count, -1, sizeof(count));

    for(int i = 0; i < order.size(); i++)
        count[order[i]] = 0;

    int i = 0;
    int j = s.size() - 1;

    while(i <= j)
    {
        if (count[s[i]] >= 0)
        {
            count[s[i]]++;
            i++;
        }
        else
        {
            char c = s[i];
            s[i] = s[j];
            s[j] = c;
            j--;
        }
    }

    int index = 0;

    for(int i = 0; i < order.size(); i++)
        for(int j = 0; j < count[order[i]]; j++)
            s[index++] = order[i];
}

int main()
{
    string s = "abdcdfs";
    string order = "dacg";

    Sort(s, order);

    cout << s << endl;
}