[LeetCode][42]trapping-rain-water

Content

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

 

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) 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.

Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9

 

Constraints:

• n == height.length
• 1 <= n <= 2 * 104
• 0 <= height[i] <= 105

Related Topics

栈

数组

双指针

动态规划

单调栈

👍 4616

👎 0

Solution

1. 单调栈+动态规划

Java

class Solution {

    public int trap(int[] height) {
        // n == height.length
        // 1 <= n <= 2 * 10⁴
        // 0 <= height[i] <= 10⁵
        int n = height.length;
        // 单调递减栈
        int[] stack = new int[n];
        int top = 0;
        // 累加bar
        int[] sums = new int[n];
        sums[0] = height[0];
        for (int i = 1; i < n; i++) {
            sums[i] = sums[i - 1] + height[i];
        }
        // dp[i]表示以下标i所对应的bar作为最右边的bar时，能trap多少雨水。
        int[] dp = new int[n];
        for (int i = 1; i < n; i++) {
            if (height[i - 1] >= height[i]) {
                stack[++top] = i;
                dp[i] = dp[i - 1];
            } else {
                while (top > 0 && height[stack[top]] < height[i]) {
                    top--;
                }
                dp[i] = dp[stack[top]]
                        + Math.min(height[stack[top]], height[i]) * (i - stack[top] - 1)
                        - (sums[i - 1] - sums[stack[top]]);
                if (height[stack[top]] >= height[i]) {
                    stack[++top] = i;
                } else {
                    stack[top] = i;
                }
            }
        }
        return dp[n - 1];
    }
}