检查字符串 S 是否可以通过用 count X 替换一些 X字符来压缩为 T

给定两个字符串， S和T ，其中S是普通字符串， T是压缩字符串，任务是确定压缩字符串T是否可以通过压缩字符串S来实现。

注意：压缩机制可以任意删除 X (X >= 0) 个字符，并用删除的字符数 (X) 替换它们。

例子：

Input: S = “HELLOWORLD”  T=”H2L4L1″
Output: True
Explanation: 
Replace index 1 to 2 in “HELLOWORLD” with count 2 -> “H2LOWORLD”
Replace index 4 to 7 in “HELLOWORLD” with count 4 -> “H2L4LD”
Replace index 9 in “HELLOWORLD” with count 1 -> “H2L4L1“
The resultant string is same as T. Hence String S can be compressed to T.

Input: S = “DFS”  T=”D1D”
Output: False
Explanation: We can clearly see that T is not a valid compressed string for S as compressed string T is ending with “D” which is not similar with the former string S.

做法：思路就是简单的遍历字符串，匹配字符如下：

• 比较 S 和 T 中相应索引处的字符，
• 同样跳过 T 中两个字母之间的索引计数。

插图：

Consider S = “HELLOWORLD”  T=”H2L4L1″

In compressed string, 2 is the integer between ‘H’ and ‘L’,so string S must has any two characters between ‘H’ and ‘L’, and it has (HELLOWORLD).

Then next integer is 4 between ‘L’ and ‘L’, then check whether String S has 4 characters or not, (HELLOWORLD)

And last integer is 1 after character ‘L’ and String also has 1 character(i.e ‘D’) after ‘L’. (HELLOWORLD)

So, this Compressed String is valid. 

S = “HELLOWORLD”  T=”H2L4L1“

请按照以下步骤解决问题：

• 开始同时遍历 S 和 T字符串直到它的长度。
• 检查 T[i] 是否为数字，如果不是数字则比较 S[j] 和 T[i] 如果它们不相等则直接返回 0 否则继续并将 j 增加 1 。
• 如果 T[i] 是一个数字，则将 j 增加到该数字，直到我们得到 T 字符串中的数字。
• 将 i 增加 1 直到 T 的长度。
• 如果 j 不等于 S 的长度，则返回 0。
• 如果满足所有条件，则最后返回 1。

下面是上述方法的实现：

C++

// C++ code for the above approach:
 
#include 
using namespace std;
 
// Return 1 if char c is number
boolean isnum(char c) { return (c >= 48 && c <= 57); }
 
// Function to check
// If string is compressed or not
int checkCompressed(string S, string T)
{
    int i = 0, j = 0;
 
    // Iterate till S.length()
    // And T.length
    while (j < S.length() && i < T.length()) {
        if (isnum(T[i]) == false) {
 
            // If its not equal
            // Then return 0
            if (S[j] != T[i]) {
                return 0;
            }
            j++;
        }
        else {
            int ans = T[i] - 48;
 
            // Iterate till we get number
            // In T string
            while (isnum(T[++i])) {
                ans *= 10;
                ans += T[i] - 48;
            }
            j += ans;
            i--;
        }
        i++;
    }
 
    // It j not equal to S string length
    // Then return 0
    if (j != S.length()) {
        return 0;
    }
    return 1;
}
 
// Driver code
int main()
{
    string S = "HelloWorld";
    string T = "H2l4l1";
    cout << checkCompressed(S, T) << endl;
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;
 
class GFG{
 
// Return 1 if char c is number
static boolean isnum(char c) { return (c >= 48 && c <= 57); }
 
// Function to check
// If String is compressed or not
static int checkCompressed(char[] S, char[] T)
{
    int i = 0, j = 0;
 
    // Iterate till S.length()
    // And T.length
    while (j < S.length && i < T.length) {
        if (isnum(T[i]) == false) {
 
            // If its not equal
            // Then return 0
            if (S[j] != T[i]) {
                return 0;
            }
            j++;
        }
        else {
            int ans = T[i] - 48;
 
            // Iterate till we get number
            // In T String
            while (i

---

Python3

# python3 code for the above approach:
 
# Return 1 if char c is number
def isnum(c):
    return ord(c) >= 48 and ord(c) <= 57
 
# Function to check
# If string is compressed or not
def checkCompressed(S, T):
 
    i, j = 0, 0
 
    # Iterate till S.length()
    # And T.length
    while (j < len(S) and i < len(T)):
        if (isnum(T[i]) == False):
 
                        # If its not equal
                        # Then return 0
            if (S[j] != T[i]):
                return 0
 
            j += 1
 
        else:
            ans = ord(T[i]) - 48
 
            # Iterate till we get number
            # In T string
            i += 1
            while (i < len(T) and isnum(T[i])):
                ans *= 10
                ans += T[i] - 48
                i += 1
 
            j += ans
            i -= 1
 
        i += 1
 
        # It j not equal to S string length
        # Then return 0
    if (j != len(S)):
        return 0
 
    return 1
 
# Driver code
if __name__ == "__main__":
 
    S = "HelloWorld"
    T = "H2l4l1"
    print(checkCompressed(S, T))
 
    # This code is contributed by rakeshsahni

C#

// C# code for the above approach:
using System;
 
public class GFG{
 
  // Return 1 if char c is number
  static boolean isnum(char c) { return (c >= 48 && c <= 57); }
 
  // Function to check
  // If String is compressed or not
  static int checkCompressed(char[] S, char[] T)
  {
    int i = 0, j = 0;
 
    // Iterate till S.length()
    // And T.length
    while (j < S.Length && i < T.Length) {
      if (isnum(T[i]) == false) {
 
        // If its not equal
        // Then return 0
        if (S[j] != T[i]) {
          return 0;
        }
        j++;
      }
      else {
        int ans = T[i] - 48;
 
        // Iterate till we get number
        // In T String
        while (i

---

Javascript

    // JavaScript Program of the above approach.
 
// Return 1 if char c is number
function isnum(c) { return (c>= 48 && c <= 57); }
  
// Function to check
// If string is compressed or not
function checkCompressed(S, T)
{
    let i = 0, j = 0;
  
    // Iterate till S.length()
    // And T.length
    while (j < S.length && i < T.length) {
        if (isnum(T[i].charCodeAt()) == false) {
  
            // If its not equal
            // Then return 0
            if (S[j] != T[i]) {
                return 0;
            }
            j++;
        }
        else {
            let ans = T[i].charCodeAt() - 48;
  
            // Iterate till we get number
            // In T string
            while (isnum(T[++i])) {
                ans *= 10;
                ans += T[i].charCodeAt() - 48;
            }
            j += ans;
            i--;
        }
        i++;
    }
  
    // It j not equal to S string length
    // Then return 0
    if (j != S.length) {
        return 0;
    }
    return 1;
}
 
    // Driver Code
    let S = "HelloWorld";
    let T = "H2l4l1";
    document.write( checkCompressed(S, T));
 
// This code is contributed by code_hunt.




输出

1

时间复杂度： O(max (|S|, |T|) )
辅助空间： O(1)