LeetCode：Trapping Rain Water

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/alading2009/article/details/72455185

Trapping Rain Water

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.

代码如下：

public int trap(int[] height) {
    if (height == null || height.length == 0) {
        return 0;
    }
    int left = 0, right = height.length - 1;
    int trap = 0, leftMax = 0, rightMax = 0;
    while (left < right) {
        leftMax = Math.max(leftMax, height[left]);
        rightMax = Math.max(rightMax, height[right]);
        if (height[left] < height[right]) {
            trap += leftMax - height[left++];
        } else {
            trap += rightMax - height[right--];
        }
    }
    return trap;
}

呼。。。打完收工