给定的字符串长度的S N由1，0，和X，该任务是代替X的每次出现按下列条件之后与最大频率打印的字符(“1”或“0”)：
- 如果存在邻接X左侧的字符为1，用1代替X。
- 如果存在邻接X右侧的字符是0，其中0替换X。
- 如果同时满足以上两个条件，则X保持不变。
注意：如果更换后1和0的频率相同，请打印X。
例子：
Input: S = “XX10XX10XXX1XX”
Output: 1
Explanation:
Operation 1: S = “X11001100X1XX”
Operation 2: S = “111001100X1XX”
No further replacements are possible.
Hence, the frequencies of ‘1’ and ‘0’ are 6 and 4 respectively.
Input: S = “0XXX1”
Output: X
Explanation:
Operation 1: S = “00X11”
No further replacements are possible.
Hence, the frequencies of both ‘1’ and ‘0’ are 2.
方法：可以根据以下观察结果解决给定问题：
- 位于“ 1”和“ 0”之间的所有“ X” (例如1XXX0 )都不重要，因为“ 1”和“ 0”都无法对其进行转换。
- 位于“ 0”和“ 1”之间的所有“ X” (例如0XXX1 )也没有意义，因为它将同等地贡献给1和0 。考虑任何形式为“ 0X….X1”的子字符串，然后在从字符串的开头和结尾更改第一次出现的X之后，子字符串中0和1的实际频率保持不变。
从以上观察结果可以得出结论，结果取决于以下条件：
- 原始字符串的“ 1”和“ 0”的计数。
- X的频率出现在两个连续的0 s或两个连续的1 s之间，分别是“ 0XXX0”和“ 1XXXX1” 。
- 出现在字符串开头且右端为“ 1”的连续“ X”的数量，即“ XXXX1…..” 。
- 出现在字符串且具有左端“ 0”的连续“ X”的数量，即…..0XXX 。
因此，根据上述条件计算1 s和0 s的数量，并打印结果计数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the most frequent
// character after replacing X with
// either '0' or '1' according as per
// the given conditions
void maxOccuringCharacter(string s)
{
// Store the count of 0s and
// 1s in the string S
int count0 = 0, count1 = 0;
// Count the frequency of
// 0 and 1
for (int i = 0;
i < s.length(); i++) {
// If the character is 1
if (s[i] == '1') {
count1++;
}
// If the character is 0
else if (s[i] == '0') {
count0++;
}
}
// Stores first occurence of 1
int prev = -1;
for (int i = 0;
i < s.length(); i++) {
if (s[i] == '1') {
prev = i;
break;
}
}
// Traverse the string to count
// the number of X between two
// consecutive 1s
for (int i = prev + 1;
i < s.length(); i++) {
// If the current character
// is not X
if (s[i] != 'X') {
// If the current character
// is 1, add the number of
// Xs to count1 and set
// prev to i
if (s[i] == '1') {
count1 += i - prev - 1;
prev = i;
}
// Otherwise
else {
// Find next occurence
// of 1 in the string
bool flag = true;
for (int j = i + 1;
j < s.length();
j++) {
if (s[j] == '1') {
flag = false;
prev = j;
break;
}
}
// If it is found,
// set i to prev
if (!flag) {
i = prev;
}
// Otherwise, break
// out of the loop
else {
i = s.length();
}
}
}
}
// Store the first occurence of 0
prev = -1;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '0') {
prev = i;
break;
}
}
// Repeat the same procedure to
// count the number of X between
// two consecutive 0s
for (int i = prev + 1;
i < s.length(); i++) {
// If the current character is not X
if (s[i] != 'X') {
// If the current character is 0
if (s[i] == '0') {
// Add the count of Xs to count0
count0 += i - prev - 1;
// Set prev to i
prev = i;
}
// Otherwise
else {
// Find the next occurence
// of 0 in the string
bool flag = true;
for (int j = i + 1;
j < s.length(); j++) {
if (s[j] == '0') {
prev = j;
flag = false;
break;
}
}
// If it is found,
// set i to prev
if (!flag) {
i = prev;
}
// Otherwise, break out
// of the loop
else {
i = s.length();
}
}
}
}
// Count number of X present in
// the starting of the string
// as XXXX1...
if (s[0] == 'X') {
// Store the count of X
int count = 0;
int i = 0;
while (s[i] == 'X') {
count++;
i++;
}
// Increment count1 by
// count if the condition
// is satisfied
if (s[i] == '1') {
count1 += count;
}
}
// Count the number of X
// present in the ending of
// the string as ...XXXX0
if (s[(s.length() - 1)]
== 'X') {
// Store the count of X
int count = 0;
int i = s.length() - 1;
while (s[i] == 'X') {
count++;
i--;
}
// Increment count0 by
// count if the condition
// is satisfied
if (s[i] == '0') {
count0 += count;
}
}
// If count of 1 is equal to
// count of 0, print X
if (count0 == count1)
{
cout << "X" << endl;
}
// Otherwise, if count of 1
// is greater than count of 0
else if (count0 > count1) {
cout << 0 << endl;
}
// Otherwise, print 0
else
cout << 1 << endl;
}
// Driver Code
int main()
{
string S = "XX10XX10XXX1XX";
maxOccuringCharacter(S);
}
// This code is contributed by SURENDAR_GANGWAR. Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to find the most frequent
// character after replacing X with
// either '0' or '1' according as per
// the given conditions
public static void
maxOccuringCharacter(String s)
{
// Store the count of 0s and
// 1s in the string S
int count0 = 0, count1 = 0;
// Count the frequency of
// 0 and 1
for (int i = 0;
i < s.length(); i++) {
// If the character is 1
if (s.charAt(i) == '1') {
count1++;
}
// If the character is 0
else if (s.charAt(i) == '0') {
count0++;
}
}
// Stores first occurence of 1
int prev = -1;
for (int i = 0;
i < s.length(); i++) {
if (s.charAt(i) == '1') {
prev = i;
break;
}
}
// Traverse the string to count
// the number of X between two
// consecutive 1s
for (int i = prev + 1;
i < s.length(); i++) {
// If the current character
// is not X
if (s.charAt(i) != 'X') {
// If the current character
// is 1, add the number of
// Xs to count1 and set
// prev to i
if (s.charAt(i) == '1') {
count1 += i - prev - 1;
prev = i;
}
// Otherwise
else {
// Find next occurence
// of 1 in the string
boolean flag = true;
for (int j = i + 1;
j < s.length();
j++) {
if (s.charAt(j) == '1') {
flag = false;
prev = j;
break;
}
}
// If it is found,
// set i to prev
if (!flag) {
i = prev;
}
// Otherwise, break
// out of the loop
else {
i = s.length();
}
}
}
}
// Store the first occurence of 0
prev = -1;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '0') {
prev = i;
break;
}
}
// Repeat the same procedure to
// count the number of X between
// two consecutive 0s
for (int i = prev + 1;
i < s.length(); i++) {
// If the current character is not X
if (s.charAt(i) != 'X') {
// If the current character is 0
if (s.charAt(i) == '0') {
// Add the count of Xs to count0
count0 += i - prev - 1;
// Set prev to i
prev = i;
}
// Otherwise
else {
// Find the next occurence
// of 0 in the string
boolean flag = true;
for (int j = i + 1;
j < s.length(); j++) {
if (s.charAt(j) == '0') {
prev = j;
flag = false;
break;
}
}
// If it is found,
// set i to prev
if (!flag) {
i = prev;
}
// Otherwise, break out
// of the loop
else {
i = s.length();
}
}
}
}
// Count number of X present in
// the starting of the string
// as XXXX1...
if (s.charAt(0) == 'X') {
// Store the count of X
int count = 0;
int i = 0;
while (s.charAt(i) == 'X') {
count++;
i++;
}
// Increment count1 by
// count if the condition
// is satisfied
if (s.charAt(i) == '1') {
count1 += count;
}
}
// Count the number of X
// present in the ending of
// the string as ...XXXX0
if (s.charAt(s.length() - 1)
== 'X') {
// Store the count of X
int count = 0;
int i = s.length() - 1;
while (s.charAt(i) == 'X') {
count++;
i--;
}
// Increment count0 by
// count if the condition
// is satisfied
if (s.charAt(i) == '0') {
count0 += count;
}
}
// If count of 1 is equal to
// count of 0, print X
if (count0 == count1) {
System.out.println("X");
}
// Otherwise, if count of 1
// is greater than count of 0
else if (count0 > count1) {
System.out.println(0);
}
// Otherwise, print 0
else
System.out.println(1);
}
// Driver Code
public static void main(String[] args)
{
String S = "XX10XX10XXX1XX";
maxOccuringCharacter(S);
}
}Python3
# Python program for the above approach
# Function to find the most frequent
# character after replacing X with
# either '0' or '1' according as per
# the given conditions
def maxOccuringCharacter(s):
# Store the count of 0s and
# 1s in the S
count0 = 0
count1 = 0
# Count the frequency of
# 0 and 1
for i in range(len(s)):
# If the character is 1
if (s[i] == '1') :
count1 += 1
# If the character is 0
elif (s[i] == '0') :
count0 += 1
# Stores first occurence of 1
prev = -1
for i in range(len(s)):
if (s[i] == '1') :
prev = i
break
# Traverse the to count
# the number of X between two
# consecutive 1s
for i in range(prev + 1, len(s)):
# If the current character
# is not X
if (s[i] != 'X') :
# If the current character
# is 1, add the number of
# Xs to count1 and set
# prev to i
if (s[i] == '1') :
count1 += i - prev - 1
prev = i
# Otherwise
else :
# Find next occurence
# of 1 in the string
flag = True
for j in range(i+1, len(s)):
if (s[j] == '1') :
flag = False
prev = j
break
# If it is found,
# set i to prev
if (flag == False) :
i = prev
# Otherwise, break
# out of the loop
else :
i = len(s)
# Store the first occurence of 0
prev = -1
for i in range(0, len(s)):
if (s[i] == '0') :
prev = i
break
# Repeat the same procedure to
# count the number of X between
# two consecutive 0s
for i in range(prev + 1, len(s)):
# If the current character is not X
if (s[i] != 'X') :
# If the current character is 0
if (s[i] == '0') :
# Add the count of Xs to count0
count0 += i - prev - 1
# Set prev to i
prev = i
# Otherwise
else :
# Find the next occurence
# of 0 in the string
flag = True
for j in range(i + 1, len(s)):
if (s[j] == '0') :
prev = j
flag = False
break
# If it is found,
# set i to prev
if (flag == False) :
i = prev
# Otherwise, break out
# of the loop
else :
i = len(s)
# Count number of X present in
# the starting of the string
# as XXXX1...
if (s[0] == 'X') :
# Store the count of X
count = 0
i = 0
while (s[i] == 'X') :
count += 1
i += 1
# Increment count1 by
# count if the condition
# is satisfied
if (s[i] == '1') :
count1 += count
# Count the number of X
# present in the ending of
# the as ...XXXX0
if (s[(len(s) - 1)]
== 'X') :
# Store the count of X
count = 0
i = len(s) - 1
while (s[i] == 'X') :
count += 1
i -= 1
# Increment count0 by
# count if the condition
# is satisfied
if (s[i] == '0') :
count0 += count
# If count of 1 is equal to
# count of 0, prX
if (count0 == count1) :
print("X")
# Otherwise, if count of 1
# is greater than count of 0
elif (count0 > count1) :
print( 0 )
# Otherwise, pr0
else:
print(1)
# Driver Code
S = "XX10XX10XXX1XX"
maxOccuringCharacter(S)
# This code is contributed by sanjoy_62.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to find the most frequent
// character after replacing X with
// either '0' or '1' according as per
// the given conditions
public static void maxOccuringCharacter(string s)
{
// Store the count of 0s and
// 1s in the string S
int count0 = 0, count1 = 0;
// Count the frequency of
// 0 and 1
for (int i = 0;
i < s.Length; i++) {
// If the character is 1
if (s[i] == '1') {
count1++;
}
// If the character is 0
else if (s[i] == '0') {
count0++;
}
}
// Stores first occurence of 1
int prev = -1;
for (int i = 0;
i < s.Length; i++) {
if (s[i] == '1') {
prev = i;
break;
}
}
// Traverse the string to count
// the number of X between two
// consecutive 1s
for (int i = prev + 1;
i < s.Length; i++) {
// If the current character
// is not X
if (s[i] != 'X') {
// If the current character
// is 1, add the number of
// Xs to count1 and set
// prev to i
if (s[i] == '1') {
count1 += i - prev - 1;
prev = i;
}
// Otherwise
else {
// Find next occurence
// of 1 in the string
bool flag = true;
for (int j = i + 1;
j < s.Length;
j++) {
if (s[j] == '1') {
flag = false;
prev = j;
break;
}
}
// If it is found,
// set i to prev
if (!flag) {
i = prev;
}
// Otherwise, break
// out of the loop
else {
i = s.Length;
}
}
}
}
// Store the first occurence of 0
prev = -1;
for (int i = 0; i < s.Length; i++) {
if (s[i] == '0') {
prev = i;
break;
}
}
// Repeat the same procedure to
// count the number of X between
// two consecutive 0s
for (int i = prev + 1;
i < s.Length; i++) {
// If the current character is not X
if (s[i] != 'X') {
// If the current character is 0
if (s[i] == '0') {
// Add the count of Xs to count0
count0 += i - prev - 1;
// Set prev to i
prev = i;
}
// Otherwise
else {
// Find the next occurence
// of 0 in the string
bool flag = true;
for (int j = i + 1;
j < s.Length; j++) {
if (s[j] == '0') {
prev = j;
flag = false;
break;
}
}
// If it is found,
// set i to prev
if (!flag) {
i = prev;
}
// Otherwise, break out
// of the loop
else {
i = s.Length;
}
}
}
}
// Count number of X present in
// the starting of the string
// as XXXX1...
if (s[0] == 'X') {
// Store the count of X
int count = 0;
int i = 0;
while (s[i] == 'X') {
count++;
i++;
}
// Increment count1 by
// count if the condition
// is satisfied
if (s[i] == '1') {
count1 += count;
}
}
// Count the number of X
// present in the ending of
// the string as ...XXXX0
if (s[s.Length - 1]
== 'X') {
// Store the count of X
int count = 0;
int i = s.Length - 1;
while (s[i] == 'X') {
count++;
i--;
}
// Increment count0 by
// count if the condition
// is satisfied
if (s[i] == '0') {
count0 += count;
}
}
// If count of 1 is equal to
// count of 0, print X
if (count0 == count1) {
Console.WriteLine("X");
}
// Otherwise, if count of 1
// is greater than count of 0
else if (count0 > count1) {
Console.WriteLine(0);
}
// Otherwise, print 0
else
Console.WriteLine(1);
}
// Driver Code
public static void Main(string[] args)
{
string S = "XX10XX10XXX1XX";
maxOccuringCharacter(S);
}
}
// This code is contributed by AnkThon输出：
1时间复杂度： O(N)
辅助空间： O(1)