hiho_1061_beautiful_string

题目大意

    题目连接：beautiful string 
    写代码之前，考虑清楚流程，以及需要维护的变量....

实现

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stack>
#include<vector>
#include<unordered_set>
#include<unordered_map>
using namespace std;
char str[10 * 1024 * 1024 + 2];
int main(){
	int T;
	scanf("%d", &T);
	while (T--){
		int n;
		scanf("%d", &n);
		getchar();
		scanf("%s", str);		
		int i = 0;
		bool valid = false;
		int beg = 0, end = 0;
		int count_1 = 0; //前2个字符连续的个数
		int count_2 = 0; //前1个字符连续的个数
		while (beg < n && !valid){
			char cur = str[beg]; //当前字符
			while (end < n && !valid){
				if (str[end] == cur){					
					//如果前两个字符都有，且当前字符数目大于等于中间字符数目，
					//则肯定可以形成一个valid的子串
					if (count_2 && count_1 && end - beg + 1 >= count_2){
						valid = true;
						break;
					}
					end++;
				}
				else{
					if (str[end] - cur != 1){//出现了一个不连续字符，则清空count_1,count_2
						count_1 = 0;
						count_2 = 0;
					}
					else{						
						//当前字符之前的那个字符的连续个数大于count_2的个数，
						//则只能以当前字符作为新的中间字符 count_2，且
						//count_1 清空
						if (count_2 && count_2 < end - beg){
							count_1 = 0;
							count_2 = end - beg;
						}
						else{
							//更新count_1 为count_2
							//更新 count_2为 end-beg;
							count_1 = count_2;
							count_2 = end - beg;
						}						
					}
					break;
				}
			}
			beg = end;
		}
		if (valid)
			printf("YES
");
		else
			printf("NO
");
	}
	return 0;
}