Container With Most Water

标签：

题目：

　　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.

思路：

 　　用两个指针分别从数组最左边 left 和最右边 right 同时往中间扫描，可以想到，倘若会出现更大的矩形，那么必定是出现在中间且高度比 left 和 right 中较短的高，所以可以得到如下关键扫描代码：

while (*left  <= high && left < right)  ++left;
while (*right <= high && left < right)  --right;

　　其中， high 是 left 和 right 中较短的值，容积由最短的高度决定，只有两边都比最短的高的情况下，容积才有可能变大。

代码：

 1 #define min(a,b) (a) > (b)? (b) : (a)
 2 int maxArea(int* height, int heightSize) {
 3     int water = 0, *left = height, *right = left + heightSize - 1;
 4     int tmp, high;
 5     while (left < right) {
 6         high = min(*left, *right);
 7         tmp = (right - left) * high;
 8         if (water < tmp)    water = tmp;
 9         while (*left  <= high && left < right)  ++left;
10         while (*right <= high && left < right)  --right;
11     }
12     return water;
13 }

Container With Most Water

Container With Most Water

标签：

原文地址：http://www.cnblogs.com/YaolongLin/p/4856595.html