Leetcode topic [array] -11- containers with the most rainwater

Force link ：https://leetcode-cn.com/probl...
Their thinking ：

1. The problem of receiving rainwater is a classic problem in the array , Let's first point out the key algorithms ： Double pointer
2. The next step is the old way , Analyze the problem stem , From the title we can see , Because the subscript of the array can be regarded as the abscissa , The value of an array can be thought of as an ordinate , Therefore, it holds the most rain , In fact, it is the largest area that can be enclosed in the array , How to calculate the maximum area ？ It's long * high , That is, the maximum area that the array subscript and value can enclose
3. Use double pointer , The slow pointer points to the array header , The fast pointer points to the end of the array , The area is a fast pointer subscript - The product of the slow pointer subscript and the smaller of the two subscripts , Record this value , Replace this value when there is a larger value
4. Array if traversal , Naturally, the shorter ones move towards the middle , Until the two hands meet

func maxArea(height []int) int {
    n := len(height)
    l, r, max := 0, n-1, 0
    for l < r {
        m := (r - l) * min(height[l], height[r])
        if max < m {
            max = m
        }
        if height[l] < height[r] {
            l++
        } else {
            r--
        }
    }
    return max
}

func min(x, y int) int {
    if x < y {
        return x
    }
    return y
}

func max(x, y int) int {
    if x > y {
        return x
    }
    return y
}