Shulou(Shulou.com)05/31 Report--

This article mainly introduces "how to use Container". In daily operation, I believe many people have doubts about how to use Container. The editor consulted all kinds of materials and sorted out simple and easy-to-use operation methods. I hope it will be helpful to answer the doubts about "how to use Container". Next, please follow the editor to study!

That is to say, there are many vertical segments on the x-axis at point 1, 2, 2, and n, with lengths A1, A2,..., and an in turn. Find out two segments so that they and x encircle the largest area. The area formula is Min (ai, aj) X | j-I | the simplest method. The for loop twice, the time complexity is O (N squared) public int maxArea (int [] height) {int n = height.length; int maxArea = 0; int I = 0; int j = nmurl; while (iheight [j]? height [j]: height [I]) / / each time the area if (height [I] < height [j]) / / the lower side of the plank moves {ibestow;} else {Jmurmuri;} if (maxArea < temp) {maxArea = temp }} return maxArea;} at this point, the study of "how to use Container" is over. I hope to be able to solve your doubts. The collocation of theory and practice can better help you learn, go and try it! If you want to continue to learn more related knowledge, please continue to follow the website, the editor will continue to work hard to bring you more practical articles!

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