LeetCode OJ Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,

Given

[0,1,0,2,1,0,1,3,2,1,2,1]

, return

6

.



The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

思路是这样的：



我们每次只考虑一单位宽度的位置能放多少水。对于某个位置来说，我们先找到它左边的最高点（有可能就是它本身），再找到它右边的最高点（有可能是它本身），然后，这个单位上最多能放min(leftMax, rightMax) - A[i]，及两边最大值的较小值减去它的高度，道理由上图显而易见。

class Solution {
public:
    int trap(int A[], int n) {
        int * lmax = new int
;
        int * rmax = new int
;
        for (int i = 0, Max = -1; i < n; i++) lmax[i] = Max = max(A[i], Max);
        for (int i = n - 1, Max = -1; i >= 0; i--) rmax[i] = Max = max(A[i], Max);
        int ans = 0;
        for (int i = 0; i < n; i++) ans += min(lmax[i], rmax[i]) - A[i];
        delete []lmax;
        delete []rmax;
        return ans;
    }
};

后来看到了常数空间复杂度的办法，思路是一样的：

class Solution {
public:
    int trap(int A[], int n) {
        int a=0;
        int b=n-1;
        int Max=0;
        int leftmax=0;
        int rightmax=0;
        while(a<=b){
            leftmax=max(leftmax,A[a]);
            rightmax=max(rightmax,A[b]);
            if(leftmax<rightmax){
                Max+=(leftmax-A[a]);
                a++;
            }
            else{
                Max+=(rightmax-A[b]);
                b--;
            }
        }
        return Max;
    }
};