[LeetCode-interview algorithm classic-Java implementation] [011-ContainerWithMostWater (the most water)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [011-ContainerWithMostWater (the most water)], leetcode -- java
[011-ContainerWithMostWater (Maximum capacity of water )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

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.

Theme

Find two vertical lines and the containers composed of these two lines and the X axis can hold the most water.

Solutions

Use greedy Algorithms
1. first, assume that we can find the vertical line of the maximum volume as I, j (assuming I <j), then the maximum volume C = min (ai, aj) * (j-I) is obtained );
2. Let's take a look at the nature below:
①: No line on the Right of j will be higher than it! Assume that k | (j <k & ak> aj) exists, then the ak> aj, So min (ai, aj, ak) = min (ai, aj ), therefore, the Container Volume C '= min (ai, aj) * (k-I)> C, which is the most important conflict with C, so we have to prove that there will be no higher lines behind j than it;
②: Similarly, there will be no higher line on the left of I. What does this mean? If we get the candidates: Set to x and y, the new two sides with a larger volume than it must be in [x, y] In the interval and ax '> = ax, ay'> = ay;
3. Therefore, we move closer from the two ends to the middle, and update the candidate values at the same time. In the contraction interval, we first contract from the smaller side of x and y;

Code Implementation

Public class Solution {public int maxArea (int [] height) {// parameter verification if (height = null | height. length <2) {return 0;} // The maximum result of the record, int result = 0; // the vertical line int left = 0 on the left; // the vertical line int right on the right = height. length-1; while (left <right) {// calculates the current maximum value result = Math. max (result, Math. min (height [left], height [right]) * (right-left); // if the right line is high if (height [left] 

Evaluation Result

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Note Please refer to the source for reprinting [http://blog.csdn.net/derrantcm/article/details/46951851]

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