How to use C # to realize the container that holds the most water
This article introduces the relevant knowledge of "how to use C# to achieve the container that holds the most water". In the operation of actual cases, many people will encounter such a dilemma, so let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!
The container that holds the most water
Difficulty: medium
I will give you n non-negative integers, a _ 1, a _ 2, and a _ 1, where each number represents a point in coordinates (I, ai). Draw n vertical lines in the coordinates. The two endpoints of the vertical line I are (I, ai) and (I, 0), respectively. Find two of the lines so that the container composed of them and the x-axis can hold the most water.
Note: you cannot tilt the container, and the value of n is at least 2.
The vertical line in the picture represents the input array [1, 8, 6, 2, 5, 5, 4, 8, 3, 7]. In this case, the maximum value that the container can hold water (represented as the blue part) is 49.
Example:
Input: [1, 8, 6, 2, 5, 4, 8, 3, 7]
Output: 49
Answer to the question:
The first solution: violence
Public int MaxArea (int [] height) {/ / solution 1: int res = 0; for (int I = 0; I
< height.Length; i++) { for (int j = i + 1; j < height.Length; j++) { int thisArea = (j - i) * Math.Min(height[i], height[j]); res = Math.Max(res, thisArea); } } return res; } 第二种解法：双指针（左指针大于右指针，left++） public int MaxArea(int[] height) { int res = 0; int left = 0; int right = height.Length - 1; while (left < right) { int thisArea = (right - left) * Math.Min(height[left], height[right]); res = Math.Max(res, thisArea); if (height[left] >Height [right]) right--; else left++;} return res;}
The third solution: double pointer optimization (left pointer less than or equal to minimum height, left++)
Public int MaxArea (int [] height) {int res = 0; int left = 0; int right = height.Length-1; while (left < right) {int minHeight = Math.Min (height [left], height [right]); int thisArea = (right-left) * minHeight; res = Math.Max (res, thisArea); if (left < right & & height [left])
