[leetcode] 11. Container with the most water - go language solution

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One 、 Title Description

Given a length of n Array of integers for height . Yes n Vertical line , The first i The two ends of the line are (i, 0) and (i, height[i]) .

Find two of them , Make them x A container of shafts can hold the most water .

Return the maximum amount of water that the container can store .

explain ： You can't tilt the container .

Example 1：

 Input ：[1,8,6,2,5,4,8,3,7]
 Output ：49 
 explain ： The vertical line in the figure represents the input array  [1,8,6,2,5,4,8,3,7]. In this case , The container can hold water （ In blue ） The maximum value of is  49.

 Example  2：
 Input ：height = [1,1]
 Output ：1

 Tips ：
n == height.length
2 <= n <= 105
0 <= height[i] <= 104

Topic link ：https://leetcode.cn/problems/container-with-most-water/

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Two 、 Their thinking - Double finger needling

Observe intuitively ：

1. The bottom of the container must be as wide as possible , This can hold more water .
2. The higher the two walls of the container, the better .
3. The amount of water in the container depends on the shorter wall .

Start from the first intuition , We can use From wide to narrow How to search for ： Use two pointers to represent the left and right boundaries of the container , Shilling points to both ends of the array , Then gradually move to the middle to search , Until the two hands meet .

Start from the second intuition , We know that the higher the two walls, the better , Then the way to move the pointer is ： Which pointer to move ？ Which wall is shorter at present , We move the pointer to the wall forward .

Every time you move , Calculate the maximum water volume of the current container , Pay attention to ： The effective height of the container shall be subject to the shorter wall .

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3、 ... and 、 My explanation

Go The language code ：

func maxArea(height []int) int {
    
    
    i:=0
    j:=len(height)-1
    max := 0

    for i != j {
    
        if height[i]<=height[j]{
    
            if (j-i)*height[i] > max {
    
                max = (j-i)*height[i]
            }
            i++
        }else{
    
            if (j-i)*height[j] > max {
    
                max = (j-i)*height[j]
            }
            j--
        }
    }
   

    return max
}

Judge the result ：