排序数组中距离字符 K 最近的较小字符
原文:https://www . geeksforgeeks . org/最近的小字符到字符的 k-from-sorted-array/
给定字符的排序数组 arr[] 和字符 K ，任务是从给定数组中找到 ASCII 值比 K 小的字符。如果没有发现字符的 ASCII 值小于 K ，则打印 -1 。 举例:
输入: arr[] = {'e '，' g '，' t '，' y'}，K = 'u' 输出: t 解释: 从' u '开始 ASCII 值最近的较小字符是' t '。 输入: arr[] = {'e '，' g '，' t '，' y'}，K = 'a' 输出: -1 说明: 不存在 ASCII 值小于' a '的字符。
天真方法: 解决问题最简单的方法是遍历数组，找到 ASCII 值小于 K 的字符，它们的 ASCII 值之差最小。 时间复杂度: O(N) 辅助空间: O(1) 高效进场: 思路是用二分搜索法找出楼元素(最大字符刚好比关键小)。按照以下步骤解决问题:
- 在阵列上执行二分搜索法。
- 检查当前中间元素( mid )是否等于字符 K 。
- 如果是这样，那么将 start 设置为mid–1，因为我们只需要找到较小的元素。
- 如果arr【mid】比 K 小，则将其存储在某个变量中，比如 ch 。将开始设置为中间+ 1 以搜索更靠近 K 的较小字符。
- 否则，将终点设置为1 中。
- 继续重复以上步骤。最后，打印完成搜索后获得的字符。如果没有得到字符，打印 -1 。
以下是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// nearest smaller character
char bs(char ar[], int n, int ele)
{
int start = 0;
int end = n - 1;
// Stores the nearest smaller
// character
char ch = '@';
// Iterate till starts cross end
while (start <= end) {
// Find the mid element
int mid = start + (end - start) / 2;
// Check if K is found
if (ar[mid] == ele)
end = mid - 1;
// Check if current character
// is less than K
else if (ar[mid] < ele) {
ch = ar[mid];
// Increment the start
start = mid + 1;
}
// Otherwise
else
// Increment end
end = mid - 1;
}
// Return the character
return ch;
}
// Driver Code
int main()
{
char ar[] = { 'e', 'g', 't', 'y' };
int n = sizeof(ar) / sizeof(ar[0]);
char K = 'u';
char ch = bs(ar, n, K);
if (ch == '@')
cout << "-1";
else
cout << ch;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to return the
// nearest smaller character
static char bs(char ar[], int n, int ele)
{
int start = 0;
int end = n - 1;
// Stores the nearest smaller
// character
char ch = '@';
// Iterate till starts cross end
while (start <= end)
{
// Find the mid element
int mid = start + (end - start) / 2;
// Check if K is found
if (ar[mid] == ele)
end = mid - 1;
// Check if current character
// is less than K
else if (ar[mid] < ele)
{
ch = ar[mid];
// Increment the start
start = mid + 1;
}
// Otherwise
else
// Increment end
end = mid - 1;
}
// Return the character
return ch;
}
// Driver Code
public static void main(String[] args)
{
char ar[] = { 'e', 'g', 't', 'y' };
int n = ar.length;
char K = 'u';
char ch = bs(ar, n, K);
if (ch == '@')
System.out.print("-1");
else
System.out.print(ch);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program to implement
# the above approach
# Function to return the
# nearest smaller character
def bs(a, n, ele):
start = 0
end = n - 1
# Stores the nearest smaller
# character
ch = '@'
# Iterate till starts cross end
while (start <= end):
# Find the mid element
mid = start + (end - start) // 2;
# Check if K is found
if(ar[mid] == ele):
end = mid - 1
# Check if current character
# is less than K
elif(ar[mid] < ele):
ch = ar[mid]
# Increment the start
start = mid + 1;
# Otherwise
else:
# Increment end
end = mid - 1;
# Return the character
return ch
# Driver code
if __name__=='__main__':
ar = [ 'e', 'g', 't', 'y' ]
n = len(ar)
K = 'u';
ch = bs(ar, n, K);
if (ch == '@'):
print('-1')
else:
print(ch)
# This code is contributed by rutvik_56
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to return the
// nearest smaller character
static char bs(char []ar, int n, int ele)
{
int start = 0;
int end = n - 1;
// Stores the nearest smaller
// character
char ch = '@';
// Iterate till starts cross end
while (start <= end)
{
// Find the mid element
int mid = start + (end - start) / 2;
// Check if K is found
if (ar[mid] == ele)
end = mid - 1;
// Check if current character
// is less than K
else if (ar[mid] < ele)
{
ch = ar[mid];
// Increment the start
start = mid + 1;
}
// Otherwise
else
// Increment end
end = mid - 1;
}
// Return the character
return ch;
}
// Driver Code
public static void Main(String[] args)
{
char []ar = { 'e', 'g', 't', 'y' };
int n = ar.Length;
char K = 'u';
char ch = bs(ar, n, K);
if (ch == '@')
Console.Write("-1");
else
Console.Write(ch);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program for the above approach
// Function to return the
// nearest smaller letacter
function bs(ar, n, ele)
{
let start = 0;
let end = n - 1;
// Stores the nearest smaller
// letacter
let ch = '@';
// Iterate till starts cross end
while (start <= end)
{
// Find the mid element
let mid = start + Math.floor((end - start) / 2);
// Check if K is found
if (ar[mid] == ele)
end = mid - 1;
// Check if current letacter
// is less than K
else if (ar[mid] < ele)
{
ch = ar[mid];
// Increment the start
start = mid + 1;
}
// Otherwise
else
// Increment end
end = mid - 1;
}
// Return the letacter
return ch;
}
// Driver Code
let ar = [ 'e', 'g', 't', 'y' ];
let n = ar.length;
let K = 'u';
let ch = bs(ar, n, K);
if (ch == '@')
document.write("-1");
else
document.write(ch);
// This code is contributed by sanjoy_62.
</script>
**Output:
t**
*时间复杂度: O(logN) 辅助空间: O(1)***
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