Interview Street Challenges - Candies

链接：https://www.interviewstreet.com/challenges/dashboard/#problem/4fe12e4cbb829

分析：可以把rating看成一个有升有降的数组，然后数组中存在三种区域（连续升、连续降、水平），区域的最小长度为1。举个例子，rating[] = [1, 2, 3, 3, 4, 2]，则direction[] = [↑, ↑, 〓, ↑, ↓]。对于每个连续区间，都从1开始计数（连续升的，从1开始递增；连续降的看成从右向左的连续升，相等的始终为1），并且记录左右边界值。以前面例子，可以分成[1, 2, 3] {left = 1, right = 3}、[1, 1] {left = 1, right = 1}、[1, 2] {left = 1, right = 2}、[2, 1] {left = 2, right = 1}这几个区间。

      现在如何利用区间数据来计算Candies呢？

section	1	2	3	1	2
section			1	1	2	1
rating	1	2	3	3	4	2

这样看来，只需在区间交错位置取最大值即可，其它直接累加区间值，最终结果就是Candies数。此例中Candies = 1 + 2 + 3 (max(3, 1)) + 1 (max(1, 1)) + 2 (max(2, 2)) + 1 =  1 + 2 + 3 + 1 + 2 + 1 = 10。

代码（AC）：

#include <cstdio>

#include <memory.h>

using namespace std;

int max(int a, int b)

{

    return a > b ? a : b;

}

// Calculate the section candies without boundary value,

// the boundary value will be set to left and right.

int continual_section_candies(int *rating, int begin, int end, int *left, int *right)

{

    int sum = 0;

    if (rating[begin] < rating[end])

    {

        *left = 1;

        *right = end - begin + 1;

        sum += (*left + *right) * (*right) / 2;

        sum -= *left;

        sum -= *right;

    }

    else if (rating[begin] > rating[end])

    {

        *right = 1;

        *left = end - begin + 1;

        sum += (*left + *right) * (*left) / 2;

        sum -= *left;

        sum -= *right;

    }

    else

    {

        sum += (end - begin - 1);

        *left = 1;

        *right = 1;

    }

    return sum;

}

int get_direction(int left, int right)

{

    if (left < right)

        return 1;

    if (left > right)

        return -1;

    return 0;

}

int main(int argc, char **argv)

{

    int rating[100005];

    int n = 0;

    // Get n.

    scanf("%d", &n);

    // Handle special case.

    if (n <= 0)

    {

        printf("0");

        return 0;

    }

    else if (n == 1)

    {

        printf("1");

        return 0;

    }

    // Init and read rating.

    memset(rating, 0, sizeof(rating));

    for (int i = 0; i < n; ++i)

    {

        scanf("%d", rating + i);

    }

    // Add a special value in rating end to 

    // make sure the rating end will be detected as a section end.

    rating[n] = rating[n - 1] == rating[n - 2] ?

        rating[n - 1] + 1 : rating[n - 2];

    // Update n.

    ++n;

    int candies = 0;                

    int last_inflection_pos = 0;   

    int last_inflection_value = 0;

    for (int i = 1; i < n - 1; ++i)

    {

        const int left_direction = get_direction(rating[i - 1], rating[i]);

        const int right_direction = get_direction(rating[i], rating[i + 1]);

        if (left_direction != right_direction)

        {

            int left, right;

            candies += continual_section_candies(rating, last_inflection_pos, i, &left, &right);

            candies += max(left, last_inflection_value);

            last_inflection_pos = i;

            last_inflection_value = right;

        }

    }

    candies += last_inflection_value;

    printf("%d", candies);

    return 0;

}