Trapping Rain Water

最新推荐文章于 2024-09-17 16:23:14 发布

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本文介绍了一个算法，用于计算给定高度数组表示的雨后能存留的雨水量。通过遍历数组并计算每个柱子两侧最高柱子的最小值，可以得出每个位置能存留的雨水量，最终累加得到总存留水量。

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.

public class Solution {
    public int trap(int[] A) {
        if (A == null || A.length <= 2) return 0;
        int maxIndex = 0;
        for (int i = 1; i < A.length; i++) {
            if (A[i] > A[maxIndex]) {
                maxIndex = i;
            }
        }
        int leftMax = A[0];
        int total = 0;
        for (int i = 1; i < maxIndex; i++) {
            if (A[i] < leftMax) {
                total += (leftMax - A[i]);
            } else {
                leftMax = A[i];
            }
        }
        int rightMax = A[A.length - 1];
        for (int i = A.length - 2; i > maxIndex; i--) {
            if (A[i] < rightMax) {
                total += (rightMax - A[i]);
            } else {
                rightMax = A[i];
            }
        }
        return total;
    }
}