LeetCode 42 Trapping Rain Water（积水体积）

题目链接： https://leetcode.com/problems/trapping-rain-water/?tab=Description

 

Problem: 根据所给数组的值，按照上图的示意图。求解积水最大体积。

 

 

首先对所给数组进行遍历操作，求出最大高度所对应的下标为maxIndex

 

之后从左向右进行遍历，设置左边高度为leftMax 并初始化为 height[0],从i==1到i==maxIndex进行遍历。不断更新water，以及leftMax

    当height[I]小于leftMax时，water += leftMax - height

    当height[i]大于leftMax时，leftMax = height[I]

 

之后从i==height.length - 2到下标 I == maxIndex进行遍历操作 初始化rightMax = height[height.length-1]

    当height[I]小于rightMax时，water+=rightMax - height[I]

    当height[I]大于rightMax时，rightMax = height[I]

 

函数结果返回water

 

参考代码：

 

package leetcode_50;


/***
 * 
 * @author pengfei_zheng
 * 求积水的体积
 */
public class Solution42 {
    public static int trap(int[] height) {
        if (height.length <= 2) return 0;
        int max = -1;
        int maxIndex = 0;
        for (int i = 0; i < height.length; i++) {
            if (height[i] > max) {
                max = height[i];
                maxIndex = i;
            }
        }
        
        int leftMax = height[0];
        int water = 0;
        for (int i = 1; i < maxIndex; i++) {
            if (height[i] > leftMax) {
                leftMax = height[i];
            } else {
                water += leftMax - height[i];
            }
        }
        
        int rightMax = height[height.length - 1];
        for (int i= height.length - 2; i > maxIndex; i--) {
            if (height[i] > rightMax) {
                rightMax = height[i];
            } else {
                water += rightMax - height[i];
            }
        }
        
        return water;
    }
    public static void main(String[]args){
//        int []height={0,1,0,2,1,0,1,3,2,1,2,1};
        int []height={10,0,11,0,10};
        System.out.println(trap(height));
    }
}