使用C ++使两个字符串相等所需的最小给定操作数。
问题陈述
给定两个字符串str1和str2，两个字符串都包含字符'a'和'b'。两个字符串的长度相等。两个字符串中都有一个_（空白）。任务是通过执行以下最少数量的操作来将第一个字符串转换为第二个字符串-
如果_在位置I上，则_可以与位置i+1或i-1处的字符交换
如果i+1和i+2位置上的字符不同，则_可以与i+1或i+2位置上的字符交换
同样，如果位置i-1和i-2处的字符不同，则_可以与位置i-1或i-2处的字符交换
如果str1=“aba_a”和str2=“_baaa”，那么我们需要2步将str1转换为str2-
1. str1 = “ab_aa” (Swapped str1[2] with str1[3]) 2. str2 = “_baaa” (Swapped str1[0] with str1[2])
算法
1. Apply a simple Breadth First Search over the string and an element of the queue used for BFS will contain the pair str, pos where pos is the position of _ in the string str. 2. Also maintain a map namely ‘vis’ which will store the string as key and the minimum moves to get to the string as value. 3. For every string str from the queue, generate a new string tmp based on the four conditions given and update the vis map as vis[tmp] = vis[str] + 1. 4. Repeat the above steps until the queue is empty or the required string is generated i.e. tmp == B 5. If the required string is generated, then return vis[str] + 1 which is the minimum number of operations required to change A to B.
示例
#include <iostream>
#include <string>
#include <unordered_map>
#include <queue>
using namespace std;
int transformString(string str, string f){
unordered_map<string, int> vis;
int n;
n = str.length();
int pos = 0;
for (int i = 0; i < str.length(); i++) {
if (str[i] == '_') {
pos = i;
break;
}
}
queue<pair<string, int> > q;
q.push({ str, pos });
vis[str] = 0;
while (!q.empty()) {
string ss = q.front().first;
int pp = q.front().second;
int dist = vis[ss];
q.pop();
if (pp > 0) {
swap(ss[pp], ss[pp - 1]);
if (!vis.count(ss)) {
if (ss == f) {
return dist + 1;
break;
}
vis[ss] = dist + 1;
q.push({ ss, pp - 1 });
}
swap(ss[pp], ss[pp - 1]);
}
if (pp < n - 1) {
swap(ss[pp], ss[pp + 1]);
if (!vis.count(ss)) {
if (ss == f) {
return dist + 1;
break;
}
vis[ss] = dist + 1;
q.push({ ss, pp + 1 });
}
swap(ss[pp], ss[pp + 1]);
}
if (pp > 1 && ss[pp - 1] != ss[pp - 2]) {
swap(ss[pp], ss[pp - 2]);
if (!vis.count(ss)) {
if (ss == f) {
return dist + 1;
break;
}
vis[ss] = dist + 1;
q.push({ ss, pp - 2 });
}
swap(ss[pp], ss[pp - 2]);
}
if (pp < n - 2 && ss[pp + 1] != ss[pp + 2]) {
swap(ss[pp], ss[pp + 2]);
if (!vis.count(ss)) {
if (ss == f) {
return dist + 1;
break;
}
vis[ss] = dist + 1;
q.push({ ss, pp + 2 });
}
swap(ss[pp], ss[pp + 2]);
}
}
return 0;
}
int main(){
string str1 = "aba_a";
string str2 = "_baaa";
cout << "Minimum required moves: " << transformString(str1, str2) << endl;
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
Minimum required moves: 2