String problem (i)

string Problems

1. Left spin problem

2. Character Inclusion issues

3. Character Matching KMP

4. Editing distance

5. Maximum palindrome string, common substring

6. Maximum common sub-sequence, Palindrome subsequence, ascending sub-sequence

7. Basic String Function implementation

8. addition, subtraction, multiplication, addition, modulus of large integers

9. Legal Palindrome, digital string

10. Regular match, longest common prefix, simplified routing

1) left-handed string

Defines the left rotation of a string: moves several characters in front of the string to the end of the string, such as ABCdef the string to the left 2 bits to get the string cdefab. A function that implements the left rotation of the string requires that the time complexity of the string operation of length n is O (n) and that the spatial complexity is O (1).

idea One, violent displacement law

Voidleftshiftone (char *s,int N) {

Char t = s[0];

Save first character

for (int i = 1; I <n; ++i) {

S[I-1] = s[i];

}

S[n-1] = t;

}

So, if you move the M-bit left, you can do the following:

void Leftshift (Char*s,int n,int m) {

while (m--) {

Leftshiftone (S,n);

}

}

thinking Second, the hand-flipping method

#include<iostream>

#include<string>

using namespacestd;

void Rotate (STRING&STR,intm) {

if (str.length () = = 0 | | m <= 0)

return;

intn = str.length ();

if (m% n <= 0)

return;

intp1 = 0, p2 = m;

intk = (n-m)-n% m;

Swap the element that p1,p2 points to, and then move the P1,P2

while (k--) {

Swap (STR[P1],STR[P2]);

p1++;

p2++;

}

Focus, all in the following lines.

Handle tail, R is the number of trailing left shift

intr = n-p2;

while (r--) {

inti = p2;

while (i> p1)

{

Swap (str[i],str[i-1]);

i--;

}

p2++;

p1++;

}

Like an example, Abcdefghijk

P1p2

When executed here, Defghia b c J k

P2+m out of Bounds,

r=n-p2=2, so the following procedure, to execute the loop two times.

First time: J step forward, Abcjk->abjck->ajbck->jabck

Then, p1++,p2++,p1 refers to A,p2 K.

P1 P2

Second time: Defghij a b C K

Likewise, after that, K moves forward, abck->abkc->akbc->kabc.

}

A few different processing optimizations were made in the tail processing

void Rotate (STRING&STR,intm) {

if (str.length () = = 0 | | m < 0)

return;

Initialize P1,P2

intp1 = 0, p2 = m;

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