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.
Note: You may not slant the container and n is at least 2.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [1,8,6,2,5,4,8,3,7] Output: 49
十分钟尝试
关键是抽象出数学模型，我没有抽象出来，看了提示后，写了一下，代码如下，二层循环
class Solution {
public int maxArea(int[] height) {
int maxArea=0;
for(int i=0;i<height.length;i++){
for(int j=i+1;j<height.length;j++){
maxArea=Math.max(maxArea,Math.min(height[i],height[j])*(j-i));
}
}
return maxArea;
}
}答案说有线性时间的解法，我尝试一下。两层for循环的线性修改就是用两个指针，代码如下：
class Solution {
public int maxArea(int[] height) {
int p=0;
int q=height.length-1;
int maxArea=0;
while(p<q){
maxArea=Math.max(maxArea,Math.min(height[p],height[q])*(q-p));
p++;
q--;
}
return maxArea;
}
}结果不对，问题出在哪里？不要两个指针都移动，这样不能全部覆盖。那个低移动哪个。
class Solution {
public int maxArea(int[] height) {
int p=0;
int q=height.length-1;
int maxArea=0;
while(p<q){
maxArea=Math.max(maxArea,Math.min(height[p],height[q])*(q-p));
if(height[p]<height[q]){
p++;
}
else{
q--;
}
}
return maxArea;
}
}