将k个断点放入数字后的最大段值
给定一个大数作为字符串s 和一个整数 k，它表示我们必须放入数字 k <=字符串长度的断点数。任务是在恰好放置 k 个断点后找到最大段值。
例子：
Input : s = "8754", k = 2
Output : Maximum number = 87
Explanation : We need to two breakpoints. After
putting the breakpoints, we get following options
8 75 4
87 5 4
The maximum segment value is 87.
Input : s = "999", k = 1
Output : Maximum Segment Value = 99
Explanation : We need to one breakpoint. After
putting the breakpoint, we either get 99,9 or
9,99.
一个重要的观察是，最大值总是长度为“string-length - k”，这是任何段的最大值。考虑到这一事实，问题变得像滑动窗口问题意味着我们需要找到所有大小子串的最大值（字符串长度 - k）。
C++
// CPP program to find the maximum segment
// value after putting k breaks.
#include
using namespace std;
// Function to Find Maximum Number
int findMaxSegment(string &s, int k) {
// Maximum segment length
int seg_len = s.length() - k;
// Find value of first segment of seg_len
int res = 0;
for (int i=0; iJava// Java program to find the maximum segment
// value after putting k breaks.
class GFG {
// Function to Find Maximum Number
static int findMaxSegment(String s, int k) {
// Maximum segment length
int seg_len = s.length() - k;
// Find value of first segment of seg_len
int res = 0;
for (int i = 0; i < seg_len; i++)
res = res * 10 + (s.charAt(i) - '0');
// Find value of remaining segments using
// sliding window
int seg_len_pow = (int)Math.pow(10,
seg_len - 1);
int curr_val = res;
for (int i = 1;
i <= (s.length() - seg_len); i++) {
// To find value of current segment,
// first remove leading digit from
// previous value
curr_val = curr_val -
(s.charAt(i - 1) - '0') * seg_len_pow;
// Then add trailing digit
curr_val = curr_val * 10 +
(s.charAt(i + seg_len - 1) - '0');
res = Math.max(res, curr_val);
}
return res;
}
// Driver code
public static void main(String[] args) {
String s = "8754";
int k = 2;
System.out.print("Maximum number = "
+ findMaxSegment(s, k));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to find the maximum segment
# value after putting k breaks.
# Function to Find Maximum Number
def findMaxSegment(s, k):
# Maximum segment length
seg_len = len(s) - k
# Find value of first segment of seg_len
res = 0
for i in range(seg_len):
res = res * 10 + (ord(s[i]) - ord('0'))
# Find value of remaining segments
# using sliding window
seg_len_pow = pow(10, seg_len - 1)
curr_val = res
for i in range(1, len(s) - seg_len):
# To find value of current segment,
# first remove leading digit from
# previous value
curr_val = curr_val - (ord(s[i - 1])-
ord('0')) * seg_len_pow
# Then add trailing digit
curr_val = (curr_val * 10 +
(ord(s[i + seg_len - 1]) - ord('0')))
res = max(res, curr_val)
return res
# Driver Code
if __name__ == '__main__':
s = "8754"
k = 2
print("Maximum number = ",
findMaxSegment(s, k))
# This code is contributed by PranchalK
C#
// C# program to find the maximum segment
// value after putting k breaks.
using System;
class GFG {
// Function to Find Maximum Number
static int findMaxSegment(string s, int k) {
// Maximum segment length
int seg_len = s.Length - k;
// Find value of first segment of seg_len
int res = 0;
for (int i = 0; i < seg_len; i++)
res = res * 10 + (s[i] - '0');
// Find value of remaining segments using
// sliding window
int seg_len_pow = (int)Math.Pow(10,
seg_len - 1);
int curr_val = res;
for (int i = 1;
i <= (s.Length- seg_len); i++) {
// To find value of current segment,
// first remove leading digit from
// previous value
curr_val = curr_val -
(s[i - 1] - '0') * seg_len_pow;
// Then add trailing digit
curr_val = curr_val * 10 +
(s[i + seg_len - 1] - '0');
res = Math.Max(res, curr_val);
}
return res;
}
// Driver code
public static void Main() {
String s = "8754";
int k = 2;
Console.WriteLine("Maximum number = "
+ findMaxSegment(s, k));
}
}
// This code is contributed by vt_m.
输出：
Maximum number = 87