【TP SRM 703 div2 250】AlternatingString

Problem Statement

A string of zeros and ones is called an alternating string if no two adjacent characters are the same. Examples of alternating strings: “1”, “10101”, “0101010101”. You are given a string s. Each character of s is a ‘0’ or a ‘1’. Please find the longest contiguous substring of s that is an alternating string. Return the length of that substring.

Definition

Class:

AlternatingString

Method:

maxLength

Parameters:

string

Returns:

int

Method signature:

int maxLength(string s)

(be sure your method is public)

Limits

Time limit (s):

2.000

Memory limit (MB):

256

Stack limit (MB):

256

Constraints

s will contain between 1 and 50 characters, inclusive.

Each character in s will be ‘0’ or ‘1’.

Examples

0)

“111101111”

Returns: 3

Among all substrings, there are 5 different alternating strings: “1”, “0”, “10”, “01”, “101”. The one with maximal length is “101” and the length is 3.

1)

“1010101”

Returns: 7

The string s itself is an alternating string.

2)

“000011110000”

Returns: 2

Note that a substring must be contiguous. The longest alternating substrings of this s are “01” and “10”. The string “010” is not a substring of this s.

3)

“1011011110101010010101”

Returns: 8

4)

“0”

Returns: 1

【题目链接】:

【题解】



找连续的01串。找长度最长的那个长度为多少.

枚举下就可以了;

应该也有O(n)的方法.

就是直接顺序往前找就可以了.

如果遇到了停顿的地方就尝试更新一下最大值并且让len=0;

因为看到长度最大为50就写了个暴力



【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);

class AlternatingString
{
    public:
    int maxLength(string s)
    {
        int len = s.size();
        int ans = 0;
        rep1(i,0,len-1)
            {
                int j = i+1;
                while (j<=len-1 && s[j]!=s[j-1])
                    j++;
                ans = max(ans,j-i);
            }
        return ans;
    }
};