Leetcode 42. 接雨水

题目传送地址: ​​https://leetcode.cn/problems/trapping-rain-water/​​

解题思路：

从中间一分为二，先算左边能接到多少雨水，然后再计算右边能接到多少雨水。最后加起来就是总和

代码如下

public static int trap(int[] height) {
        //该位置往右的方向，最高峰的高度是多少
        int result = 0; //总雨水数
        int left = 0, right = 1;
        // 先算左侧接到的雨水
        for (int i = 0; i < height.length; i++) {
            if (right == 1) {
                while (right < height.length) {
                    if (height[left] > height[right]) {
                        right++;
                    } else {
                        break;
                    }
                }
                if (right==height.length){
                    break;
                }
            }
            if (i < right) {
                result += height[left] - height[i];
            }
            if (i == right) {
                //那就得找下一个右高峰
                left = right;
                right++;
                while (right < height.length) {
                    if (height[left] > height[right]) {
                        right++;
                    } else {
                        break;
                    }
                }
                if (right == height.length) {
                    break;
                }
            }
        }
        
        //再算右边接到的雨水
        right = height.length - 1;
        for (int i = height.length - 1; i > left; i--) {
            if (height[i] < height[right]) {
                result += height[right] - height[i];
            }
            if (height[i] >= height[right]) {
                right = i;
            }
        }
        return result;
    }