[LeetCode] Container With Most Water

Container With Most Water

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.

解题思路：

这道题的意思是直角平面上有n条垂直于x轴的线，相邻的线相距1，选取两根线，与x轴构成一容器。求这样的容器能最多够装多少水？

1、基本方法，两两组合，计算出最大值。时间复杂度为O（n^2）。产生超时错误。

class Solution {
public:
int maxArea(vector<int>& height) {
//naiv办法
int len = height.size();
if(len<2){ //至少要有两个以上的线方能组成容器
return 0;
}
int maxArea=0;
int minHeight = 0;
for(int i=1; i<len; i++){
minHeight = height[i];
for(int j=i-1; j>=0; j--){
minHeight = min(height[i], minHeight);
if(minHeight==0){
break;
}
maxArea=max(maxArea, minHeight*(i-j));
}
}
return maxArea;
}

int min(int a, int b){
return a>b?b:a;
}

int max(int a, int b){
return a>b?a:b;
}
};2、比较talent的办法，两边夹，若左边木板小于右边，移动左边木板，否则移动右边木板。因为只有这样容量才有可能变大。
class Solution {
public:
int maxArea(vector<int>& height) {
int maxArea = 0;
int len = height.size();
int i=0, j=len-1;
while(i<j){
maxArea=max(maxArea, min(height[i], height[j])*(j-i));
if(height[i]<height[j]){
i++;
}else{
j--;
}
}

return maxArea;
}

int max(int a, int b){
return a>b?a:b;
}

int min(int a, int b){
return a>b?b:a;
}
};