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 and n is at least 2.
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题目
给定 n 个非负整数 a1, a2, ..., an, 每个数代表了坐标中的一个点 (i, ai)。画 n 条垂直线，使得 i 垂直线的两个端点分别为(i, ai)和(i, 0)。找到两条线，使得其与 x 轴共同构成一个容器，以容纳最多水。
注意事项
容器不可倾斜。
样例
给出[1,3,2], 最大的储水面积是2
分析
用两根指针一头一尾，分别计算面积，然后选取高度小的那边向中间移动，因为这样才可能存在更大的面积，更新最大的面积，最后就是结果。
Paste_Image.png
代码
public class Solution {
/**
* @param heights: an array of integers
* @return: an integer
*/
int computeArea(int left, int right, int[] heights) {
return (right-left)*Math.min(heights[left], heights[right]);
}
public int maxArea(int[] heights) {
//两根指针，一头一尾计算，短的那根指针向中间移动，因为只有这种情况才存在更大的情况
int left = 0, right = heights.length-1;
int res = 0;
while(left < right) {
res = Math.max(res, computeArea(left,right,heights));
if(heights[left] < heights[right]) {
left++;
//如果小于则不必要计算，直接跳过
while(heights[left] <= heights[left-1] && left<right)
left++;
}
else {
right--;
while(heights[right] <= heights[right+1] && left <right)
right--;
}
}
return res;
}
}
