【Leetcode】Rain trapping

最新推荐文章于 2025-09-13 14:08:34 发布

转载最新推荐文章于 2025-09-13 14:08:34 发布 · 367 阅读

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本文介绍了一种高效算法，用于计算给定地形高度图在降雨后的积水总量。通过维护左右两侧的最大高度并逐步填充低洼区域，该算法实现了O(n)的时间复杂度。

【题目】

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

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1],return 6.

【思路】

Note: The following idea is in fact from the last answer in this link, which leads to a clean code. I just reorganize it and add some explanations. I hope it is Ok.

link:

Here is my idea: instead of calculating area by height*width, we can think it in a cumulative way. In other words, sum water amount of each bin(width=1). Search from left to right and maintain a max height of left and right separately, which is like a one-side wall of partial container. Fix the higher one and flow water from the lower part. For example, if current height of left is lower, we fill water in the left bin. Until left meets right, we filled the whole container.

class Solution {
public:
    int trap(int A[], int n) {
        int left=0; int right=n-1;
        int res=0;
        int maxleft=0, maxright=0;
        while(left<=right){
            if(A[left]<=A[right]){
                if(A[left]>=maxleft) maxleft=A[left];
                else res+=maxleft-A[left];
                left++;
            }
            else{
                if(A[right]>=maxright) maxright= A[right];
                else res+=maxright-A[right];
                right--;
            }
        }
        return res;
    }
};

The basic idea is that we set two pointers l and r to the left and right end of height. Then we get the minimum height (minHeight) of these pointers (similar to Container with Most Water due to the Leaking Bucket Effect) since the level of the water cannot be higher than it. Then we move the two pointers towards the center. If the coming level is less than minHeight, then it will hold some water. Fill the water until we meet some "barrier" (with height larger than minHeight) and update l and rto repeat this process in a new interval.

【代码】

public class Solution {
    public int trap(int[] height) {
        int n = height.length, l = 0, r = n - 1, water = 0, minHeight = 0;
        while (l < r) {
            while (l < r && height[l] <= minHeight)
                water += minHeight - height[l++];
            while (r > l && height[r] <= minHeight)
                water += minHeight - height[r--];
            minHeight = Math.min(height[l], height[r]);
        }
        return water;
    }
}