【LeetCode】 11. Container With Most Water
【LeetCode】 11. Container With Most Water
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本文链接：https://blog.csdn.net/shiliang97/article/details/101952213
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
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/container-with-most-water 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
暴力直接过了击败了百分之12的人
class Solution {
public:
int min(int a,int b){
if(a>b){
return b;
}return a;
}
int maxArea(vector<int>& height) {
int max=-1;
for(int i=0;i<height.size();i++){
for(int l=i+1;l<height.size();l++){
int num=min(height[i],height[l])*(l-i);
if(num>max){
max=num;
}
}
}
return max;
}
};
整了个双指针，速度时间一下提升了不少
class Solution {
public:
int maxArea(vector<int>& height) {
int a=0,b=height.size()-1;
int max=-1;
while(a!=b){
if(height[a]<=height[b]){
if(height[a]*(b-a)>max){
max=height[a]*(b-a);
}
a++;
}else{
if(height[b]*(b-a)>max){
max=height[b]*(b-a);
}
b--;
}
}
return max;
}
};看到一段特别有意义的评论
证明很严谨。 其实无论是移动短指针和长指针都是一种可行求解。 只是，一开始就已经把指针定义在两端，如果短指针不动，而把长指针向着另一端移动，两者的距离已经变小了，无论会不会遇到更高的指针，结果都只是以短的指针来进行计算。 故移动长指针是无意义的。
最快的人咋写的？
class Solution {
public:
int maxArea(vector<int>& height)
{
int maxarea = 0, l = 0, r = height.size() - 1;
while (l < r) {
int a = height[l] > height[r] ? height[r] : height[l];
maxarea = maxarea > a * (r - l) ? maxarea : a * (r - l);
if (height[l] < height[r])
l++;
else
r--;
}
return maxarea;
}
};腾讯云开发者
