贪心法,每个容器的面积取决于最短的木板,时间复杂度o(n),空间复杂度o(1)
class Solution {
public:
int maxArea(vector<int> &height) {
int start = 0;
int end = height.size() - 1;
int result = INT_MIN;
while (start < end) {
int area = min(height[end], height[start]) * (end - start);
result = max(result, area);
if (height[start] <= height[end]) {
start++;
} else {
end--;
}
}
return result;
}
};