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.
a1,a2,......an是高度,它们之间的距离是底长,容器的面积是两个高度中最短的那个边乘以底长。找这个数列中能使容器面积最大的。设两个指针,一个在最左,一个在最右,都往中间走。判断条件是:如果height[low] < height[high] 那么 low++, 否则high--。 因为如果height[low] < height[high],那么面积由low决定,这时high往左走,面积是不会增大的(不仅底长减小,而且要么high-1位置上的数大于low位置上的数面积由low决定,要么high-1位置上的数小于low位置上的数由此位决定,总之面积都是变小),所以唯一面积可能增大的方法就是low++。
Source
public int maxArea(int[] height) {
if(height.length == 0) return 0;
int max = 0;
int cap = 0;
int low = 0, high = height.length - 1;
while(low < high){
cap = (high - low) * Math.min(height[low], height[high]);
if(max < cap) max = cap;
if(height[low] < height[high]){
low ++;
}
else{
high --;
}
}
return max;
}
Test
public static void main(String[] args){
int[] height = {3,2,6,4,7,9};
int b = new Solution().maxArea(height);
System.out.println(b);
}