LeetCode-Trapping Rain Water

又是这种需要仔细考虑的题目，

很容易就会漏掉一些情况，哎，感冒了，

完全不在状态，调了好几个小时，

写代码的能力实在是不行，总感觉写出来的代码不够优雅，

class Solution {
public:
    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int res = 0;
        if (A == NULL) {
            return res;
        }
        int i = 1;
        int b = 0;
        int low;
        while (i < n && A[i] >= A[i - 1]) {
            b = i;
            ++i;
        }
        while (i < n) {
            while (i < n && A[i] < A[i - 1]) {
                ++i;
            }
            int k = i - 1;
            int j = k - 1;
            low = A[k];
            while (i < n && A[i] >= A[i - 1]) {
                if (A[i] > A[i - 1]) {
                    if (j >= b) {
                        while (j >= b && A[i] >= A[j]) {
                            int len = i - j - 1;
                            int h = A[j] - low;
                            low = A[j];
                            res += len * h;
                            while (j >= b && A[j] <= low) 
                                --j;
                        }
                        if (j >= b) {
                            int len = i - j - 1;
                            int h = A[i] - low;
                            low = A[i];
                            res += len * h;
                        }
                    }
                    ++i;
                    continue;
                }
                ++i;
            }
            if (A[b] <= A[i - 1]) {
                b = i - 1;
            }
            ++i;
        }
        return res;
    }
};

大神们的方法，超级简单，找左右两边的最大值即可；

class Solution {
public:
    int trap(int A[], int n) {
        if (n == 0) return 0;
        int left[n];
        int right[n];

        int lmax = A[0];
        for (int i = 0; i < n; ++i)
        {
            lmax = max(lmax, A[i]);
            left[i] = lmax;
        }

        int rmax = A[n-1];
        for (int i = n - 1; i >= 0; --i)
        {
            rmax = max(rmax, A[i]);
            right[i] = rmax;
        }

        int res = 0;
        for (int i = 0; i < n; ++i)
            res += min(left[i], right[i]) - A[i];

        return res;
    }
};