Largest Rectangle in Histogram

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given height = [2,1,5,6,2,3],
return 10.

思路：先新建两个数组left和right，left数组用于保存每个矩形的最左边开始的可以和当前矩形形成矩形的最小下标，

right数组用于保存每个矩形的最右边开始的可以和当前矩形形成矩形的最大下标。

最后，根据最左最右下标的差值相乘矩形的高度值，求出矩形的最大值。

class Solution {
public:
    int max (int a, int b)
{
    return a > b ? a : b;
}
int largestRectangleArea(vector<int>& height) {
    int n = height.size();
    int *left,*right,next,i;
    left = (int*)malloc(n * sizeof(int));
    right = (int*)malloc(n * sizeof(int));
    left[0] = -1;
    right[n-1] = n;
    for (i=1; i < n; i++){
        if (height[i-1] < height[i]){
            left[i] = i-1;
        }else {
            next = i-1;
            while (left[next] != -1 && height[left[next]] >= height[i])
                next = left[next];
            left[i] = left[next];
        }
    }
    for (i=n-2; i >= 0; i--){
        if (height[i+1] < height[i]){
            right[i] = i+1;
        }else {
            next = i+1;
            while (right[next] != n && height[right[next]] >= height[i])
                next = right[next];
            right[i] = right[next];
        }
    }
    int largest = 0;
    for (i=0 ;i < n; i++){
        largest = max(largest,height[i] * (right[i]-left[i]-1));
    }
    return largest;
        }
};