Container With Most Water

Container With Most Water

问题：

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container.

思路：

　　双指针，减少短板

我的代码：

public class Solution {

    public int maxArea(int[] height) {

        if(height == null || height.length <= 1)    return 0;

        int left = 0;

        int right = height.length - 1;

        int maxArea = Math.min(height[left],height[right])*(right - left);

        while(left < right)

        {

            int curLeft = left;

            int curRight = right;

            if(height[left] < height[right])

            {

                while(left < right && height[left] <= height[curLeft])

                {

                    left++;

                }

                maxArea = Math.max(Math.min(height[left],height[right])*(right - left), maxArea);

            }

            else

            {

                while(left < right && height[right] <= height[curRight])

                {

                    right--;

                }

                maxArea = Math.max(Math.min(height[left],height[right])*(right - left), maxArea);

            }

        }

        return maxArea;

    }

}

View Code

他人代码：

public class Solution {

    public int maxArea(int[] height) {

        if (height == null) {

            return 0;

        }

        

        int left = 0;

        int right = height.length - 1;

        int maxArea = 0;

        

        while (left < right) {

            int h = Math.min(height[left], height[right]);

            int area = h * (right - left);

            maxArea = Math.max(maxArea, area);

            

            if (height[left] < height[right]) {

                // 如果左边界比较低，尝试向右寻找更高的边界

                left++;

            } else {

                // 如果右边界比较低，尝试向左寻找更高的边界

                right--;

            }

        }

        

        return maxArea;

    }

}

View Code

学习之处：

• 思路一致，他人的代码更加简洁清楚，我的代码虽然循环多，但是时间复杂度是一致的，都是O(n)，另外我的代码里面少计算了好多次maxArea