原文:
给定一个字符串数组和一个整数数组，其中数组中的第 i 第t3【整数】对应于字符串数组中的第 i 第t7】字符串的值。现在选择整数数组中具有最小值的两个字符串，将两个整数相加并连接两个字符串，将相加的整数加到整数数组中，将连接的字符串加到字符串数组中，并继续重复整个过程，直到两个数组中只剩下一个元素。此外，请注意，一旦选择了任意两个字符串和整数，它们就会从数组中删除，并且新的字符串和整数会添加回各自的数组中。 任务是打印得到的最终字符串。****
示例:
输入: str = {“极客”、“for”、“极客”}，num = {2，3，7} 输出:极客 forgeeks 选择 2 和 3 添加它们，形成“极客 for”和 5 将它们添加回数组，str = {“极客 for”、“极客”}，num = {5，7} 现在选择 7 和 5 添加它们，形成“极客 forgeeks”，这是最后的字符串。
输入:str = { " abc "，" def "，" ghi "，" jkl " }，num = {1，4，2，6} 输出:jklabcghdef
方法:思路是使用一个，制作一对字符串及其对应的值，存储在优先级队列中，保持两对从优先级队列中出列，将整数相加并串接，重新排入优先级队列，直到队列的大小减为 1，然后打印队列中唯一剩余的字符串。
下面是上述方法的实现:
c
// c implementation of the approach
#include
using namespace std;
// class that represents a pair
struct priority
{
string s;
int count;
};
struct compare
{
int operator()(priority a, priority b)
{
if (a.count > b.count)
return 1;
else if (a.count < b.count)
return -1;
return 0;
}
};
// function that prints the final string
static void findstring(string str[], int num[],
int n)
{
priority_queue, compare> p;
// add all the strings and their corresponding
// values to the priority queue
for(int i = 0; i < n; i )
{
priority x;
x.s = str[i];
x.count = num[i];
p.push(x);
}
// take those two strings from the priority
// queue whose corresponding integer values
// are smaller and add them up as well as
// their values and add them back to the
// priority queue while there are more
// than a single element in the queue
while (p.size() > 1)
{
// get the minimum valued string
priority x = p.top();
p.pop();
// p.remove(x);
// get the second minimum valued string
priority y = p.top();
p.pop();
// updated integer value
int temp = x.count y.count;
string sb = x.s y.s;
// create new entry for the queue
priority z;
z.count = temp;
z.s = sb;
// add to the queue
p.push(z);
}
// print the only remaining string
priority z = p.top();
p.pop();
cout << z.s << endl;
}
// driver code
int main()
{
string str[] = { "geeks", "for", "geeks" };
int num[] = { 2, 3, 7 };
int n = sizeof(num) / sizeof(int);
findstring(str, num, n);
}
// this code is contributed by sanjeev2552
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of the approach
import java.util.*;
import java.lang.*;
// class that represents a pair
class priority {
string s;
int count;
}
class pq implements comparator {
public int compare(priority a, priority b)
{
if (a.count > b.count)
return 1;
else if (a.count < b.count)
return -1;
return 0;
}
}
class gfg {
// function that prints the final string
static void findstring(string str[], int num[], int n)
{
comparator comparator = new pq();
priorityqueue p
= new priorityqueue(comparator);
// add all the strings and their corresponding
// values to the priority queue
for (int i = 0; i < n; i ) {
priority x = new priority();
x.s = str[i];
x.count = num[i];
p.add(x);
}
// take those two strings from the priority
// queue whose corresponding integer values are smaller
// and add them up as well as their values and
// add them back to the priority queue
// while there are more than a single element in the queue
while (p.size() > 1) {
// get the minimum valued string
priority x = p.poll();
p.remove(x);
// get the second minimum valued string
priority y = p.poll();
p.remove(y);
// updated integer value
int temp = x.count y.count;
string sb = x.s y.s;
// create new entry for the queue
priority z = new priority();
z.count = temp;
z.s = sb;
// add to the queue
p.add(z);
}
// print the only remaining string
priority z = p.poll();
system.out.println(z.s);
}
// driver code
public static void main(string[] args)
{
string str[] = { "geeks", "for", "geeks" };
int num[] = { 2, 3, 7 };
int n = num.length;
findstring(str, num, n);
}
}
java 描述语言
output:
geeksforgeeks
时间复杂度: o(n * log(n)) 辅助空间: o(n)
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