编程算法

扑克牌的顺子 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目: 从扑克牌中随机抽取5张牌, 推断是不是一个顺子, 即这5张牌是不是连续的.

2~10为数字本身, A为1, J为11, Q为12, K为13, 而大小王能够看成随意数字.

排序, 推断字符串之间的间隔数, 假设小于等于大小王的数量, 则是连续, 否则不是.

代码:

/*
 * main.cpp
 *
 *  Created on: 2014.7.12
 *      Author: spike
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

using namespace std;

int compare(const void* num1, const void* num2) {
	return *(int*)num1-*(int*)num2;
}

bool IsContinuous(int* numbers, int length) {
	if (numbers == NULL || length < 1)
		return false;
	qsort(numbers, length, sizeof(int), compare);
	int numberOfZero = 0;
	int numberOfGap = 0;
	for (int i=0; i<length&&numbers[i]==0; ++i)
		++numberOfZero;
	int small = numberOfZero;
	int big = small+1;
	while(big < length) {
		if (numbers[small] == numbers[big])
			return false;
		numberOfGap += numbers[big] - numbers[small]-1;
		small = big;
		++big;
	}
	return (numberOfGap>numberOfZero) ? false : true;
}

int main(void)
{
    int numbers[] = {0, 3, 1, 4, 5};
    bool result = IsContinuous(numbers, 5);
    printf("result = %s
", result ? "true" : "false");
    return 0;
}

输出:

result = true