LeetCode_11---Container With Most Water

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.

翻译：就是说，x轴上在1,2,...,n点上有许多垂直的线段，长度依次是a₁, a₂, ..., an。找出两条线段，使他们和x抽围成的面积最大。面积公式是 Min(a_i, a_j) X |j - i|

Code:

import java.util.HashMap;
import java.util.Map;

/**
 * 
 */

/**
 * @author MohnSnow
 * @time 2015年6月2日 下午5:24:20
 * 
 */
public class LeetCode11 {

	/**
	 * @param argsmengdx
	 *            -fnst
	 */
	//梯形面积=（上底+下底）*高/2
	//Max = (aj+ai)*(j-i)/2
	//Time Limit Exceeded---->可见这个空间复杂度有点高，可以尝试利用空间复杂度取代时间复杂度
	//一开始没看懂题目，以为是一个梯形了
	public static int maxArea(int[] height) {
		int len = height.length;
		//System.out.println("maxArea: len:" + len);
		int max = 0;
		int temp = 0;
		int i = 0, j = 0;
		for (; i < len; i++) {
			for (j = i + 1; j < len; j++) {
				temp = (Math.min(height[i], height[j])) * (j - i);
				if (max < temp) {
					//System.out.println("maxArea: i:" + i + " j:" + j);
					max = temp;
				}
			}
		}
		return max;
	}

	//直观的解释是：容积即面积，它受长和高的影响，当长度减小时候，高必须增长才有可能提升面积
	//所以我们从长度最长时开始递减，然后寻找更高的线来更新候补；
	//还是从两头向中间靠拢的解法
	public static int maxArea1(int[] height) {
		int len = height.length;
		int max = 0;
		int i = 0, j = len - 1;
		while (i < j) {
			max = Math.max(Math.min(height[i], height[j]) * (j - i), max);
			if (height[i] < height[j]) {
				int m = i + 1;
				while (m < j && height[m] < height[i]) {
					m++;
				}
				i = m;
			} else {
				int m = j - 1;
				while (m > i && height[m] < height[j]) {
					m--;
				}
				j = m;
			}
		}
		System.out.println("maxArea1: i:" + i + " j:" + j);
		return max;
	}

	public static int maxArea2(int[] height) {
		int i = 0, j = height.length - 1;
		int max = 0;
		while (i < j) {
			max = Math.max(Math.min(height[i], height[j]) * (j - i), max);
			if (height[i] < height[j]) {
				i++;
			} else {
				j--;
			}
		}
		return max;
	}

	/**
	 * @param i
	 * @param j
	 * @returnmengdx-fnst
	 */
	public static void main(String[] args) {
		//int[] test = { 1, 2, 3, 4, 8, 9, 6, 5, 4, 9, 6, 23, 48, 5, 6, 56, 5, 46, 6 };
		int[] test = { 1, 2, 5, 4 };
		System.out.println(maxArea(test));
		System.out.println(maxArea1(test));
		System.out.println(maxArea2(test));
	}

}