给定一个长度为N的字符串S ，由小写英文字母和“?”组成，任务是检查是否有可能通过替换所有的‘?’使字符串不减少。以英文字母字符。如果发现是真的，打印“是” 。否则，打印“否” 。
例子：
Input: S = “abb?xy?”
Output: Yes
Explanation:
Replacing the ‘?’s at index 3 and 6 with ‘b’ and ‘z’, modifies the string S to “abbbxyz”, whcih is non-decreasing.
Input: S = “??z?a?”
Output : No
朴素的方法：解决问题的最简单方法是使用英文字母生成所有可能的字符串，并检查生成的字符串是否不减。如果发现是真的，打印“是” ，否则打印“否”。
时间复杂度： O(N*26 N )
辅助空间： O(N)
有效的方法：可以根据以下观察解决给定的问题：
- It can be observed that the only position that matters is the position with S[i]! = ‘?’, as one can replace the ‘?’ with the character at the nearest index that is not ‘?’.
- Therefore, the task is reduced to checking if characters of the strings that are not ‘?’, are in non-decreasing order or not.
请按照以下步骤解决问题：
- 初始化一个变量，比如last，来存储最后一个不是‘?’ 的字符。
- 遍历字符串，对于每个第i个字符，检查是否S[i]! = ‘?’和S[i]
- 否则，如果S[i]!=’?’ ，则将last更新为last = S[i] 。
- 如果以上情况都不满足，则打印“是” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if it's possible
// to make the string s non decreasing
int nonDecreasing(string s)
{
// Stores length of string
int size = s.length();
// Stores the last character
// that is not '?'
char c = 'a';
// Traverse the string S
for (int i = 0; i < size; i++) {
// If current character
// of the string is '?'
if (s[i] == '?') {
continue;
}
// If s[i] is not '?'
// and is less than C
else if ((s[i] != '?')
&& (s[i] < c)) {
return 0;
}
// Otherwise
else {
// Update C
c = s[i];
}
}
// Return 1
return 1;
}
// Driver Code
int main()
{
string S = "abb?xy?";
if (nonDecreasing(S))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to check if it's possible
// to make the String s non decreasing
static int nonDecreasing(char []s)
{
// Stores length of String
int size = s.length;
// Stores the last character
// that is not '?'
char c = 'a';
// Traverse the String S
for (int i = 0; i < size; i++)
{
// If current character
// of the String is '?'
if (s[i] == '?')
{
continue;
}
// If s[i] is not '?'
// and is less than C
else if ((s[i] != '?')
&& (s[i] < c))
{
return 0;
}
// Otherwise
else {
// Update C
c = s[i];
}
}
// Return 1
return 1;
}
// Driver Code
public static void main(String[] args)
{
String S = "abb?xy?";
if (nonDecreasing(S.toCharArray())==1)
System.out.print("Yes" +"\n");
else
System.out.print("No" +"\n");
}
}
// This code is contributed by 29AjayKumarPython3
# python 3 program for the above approach
# Function to check if it's possible
# to make the string s non decreasing
def nonDecreasing(s):
# Stores length of string
size = len(s)
# Stores the last character
# that is not '?'
c = 'a'
# Traverse the string S
for i in range(size):
# If current character
# of the string is '?'
if (s[i] == '?'):
continue
# If s[i] is not '?'
# and is less than C
elif((s[i] != '?') and (s[i] < c)):
return 0
# Otherwise
else:
# Update C
c = s[i]
# Return 1
return 1
# Driver Code
if __name__ == '__main__':
S = "abb?xy?"
if(nonDecreasing(S)):
print("Yes")
else:
print("No")
# This code is contributed by SURENDRA_GANGWAR.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to check if it's possible
// to make the String s non decreasing
static int nonDecreasing(char []s)
{
// Stores length of String
int size = s.Length;
// Stores the last character
// that is not '?'
char c = 'a';
// Traverse the String S
for (int i = 0; i < size; i++)
{
// If current character
// of the String is '?'
if (s[i] == '?')
{
continue;
}
// If s[i] is not '?'
// and is less than C
else if ((s[i] != '?')
&& (s[i] < c))
{
return 0;
}
// Otherwise
else {
// Update C
c = s[i];
}
}
// Return 1
return 1;
}
// Driver Code
public static void Main(String[] args)
{
String S = "abb?xy?";
if (nonDecreasing(S.ToCharArray())==1)
Console.Write("Yes" +"\n");
else
Console.Write("No" +"\n");
}
}
// This code is contributed by 29AjayKumar输出：
Yes时间复杂度： O(N)
辅助空间： O(1)
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