Java实现Container With Most Water

算法描述    

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.

        注意: You may not slant the container and n is at least 2.

算法一实现：

public class Solution {    /**     * Given n non-negative integers a1, a2, ..., an, where each represents<br>     * a point at coordinate (i, ai). n vertical lines are drawn such that<br>     * the two endpoints of line i is at (i, ai) and (i, 0). Find two lines,<br>      * which together with x-axis forms a container, such that the container <br>     * contains the most water.<br>     * <font color="red">Note: You may not slant the container and n is at least 2.</font>     * @param  height int[]     * @return maxArea int     */    public int maxArea(int[] height) {        //height不能为空        if (height == null) {            throw new NullPointerException("height数组不能为空");        }        //数组元素为两个        if (height.length == 2) {            if (height[0] == 0 || height[1] == 0)   return 0;            else if (height[0] < height[1])         return height[0];            else if (height[0] >= height[1])        return height[1];        }        //数组元素超过两个        if (height.length > 2) {            int maxArea  = 0;            for (int i = 0; i < height.length; i++) {                if (height[i] == 0) {                    continue;                }                int area = 0;                for (int j = i + 1; j < height.length; j++) {                    if (height[j] == 0) {                        continue;                    }                    int areaTemp = 0;                    if (height[i] < height[j]) {                        areaTemp = height[i] * (j - i);                    } else {                        areaTemp = height[j] * (j - i);                    }                    if (area < areaTemp) {                        area = areaTemp;                    }                }                if (maxArea < area) {                    maxArea = area;                }            }            return maxArea;        }        //数组元素个数小于2时，值为0        return 0;    }}

缺点：算法时间复杂度高，效率低。

算法二实现（优化）

public class Solution {    public int maxArea(int[] height) {        //height不能为空        if (height == null) {            throw new NullPointerException("height数组不能为空");        }                //数组个数不小于2时        if (height.length >= 2) {            int maxArea = 0;            int i = 0;            int j = height.length - 1;            while (i != j) {                int area = 0;                if (height[i] < height[j]) {                    area = height[i] * (j - i);                    i++;                } else {                    area = height[j] * (j - i);                    j--;                }                if (maxArea < area) {                    maxArea = area;                }            }            return maxArea;        }        //数组元素个数小于2时，值为0        return 0;    }}

变成for循环

public int maxArea(int[] height) {        //height不能为空        if (height == null) {            throw new NullPointerException("height数组不能为空");        }                //数组个数不小于2时        if (height.length >= 2) {            int maxArea = 0;            for (int i = 0, j = height.length - 1, area = 0; i != j; ) {                if (height[i] < height[j]) {                    area = height[i] * (j - i);                    i++;                } else {                    area = height[j] * (j - i);                    j--;                }                if (maxArea < area) {                    maxArea = area;                }            }            return maxArea;        }        //数组元素个数小于2时，值为0        return 0;    }

算法二证明：

      在a1，a2.....an中，存在两个点使所求面积最大，所以假设ax，ay为其中使面积最大的两个点，i < j。我们需要证明的是一定能遍历到这两个点即可。i从左到右移动，j从右往左。i和j对应的值哪个小哪个移动，这样始终在找大值，越移动j - i越来越小，高度越来越大，所以一定能找到ax和ay。

再具体证明一下，如下图所示：

图1

        我们需要证明两点：一.要想使ax和ay所在面积最大，a（y+1）的值一定是1或2的位置（即：比ax小的位置）；二：如何保证a(y+1)移动到a（y）。

i当前在ax的位置，j当前在a（y+1）的位置。当j向左移动后，j-i变小了，如果想使ax和ay所在面积最大，只能是a（y+1）的值在1和2的位置；要使j向左移动，则只要height[i] > height[j]。所以一定可以移动到ay的位置，证明完成。