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.
题解:
暴力解即可,注意这个面积是长方形的,我刚开始犯傻给写成梯形了。。。两个思路:两个for循环遍历,结果就是TLE了。另一个就是头尾指针遍历法,时间复杂度降为O(n).以后遇到这种需要遍历的,要先想到头尾指针法,而不是两个for循环(其实也是双指针,只是位置不一样)。同时,与之前遇到的数组问题一样,小幅度的优化就是在大循环内就跳过重复项
Solution 1 (TLE)
class Solution {
public:
int maxArea(vector<int>& height) {
int area = , n = height.size();
for(int i=; i<n-; i++) {
for(int j=i+; j<n; j++) {
int new_area = (j-i)*min(height[i], height[j]);
if(height[i]== || height[j]==) new_area = ;
area = max(area, new_area);
}
}
return area;
}
};
Solution 2
class Solution {
public:
int maxArea(vector<int>& height) {
int area = , i = , j = height.size() - ;
while (i < j) {
area = max(area, min(height[i], height[j]) * (j - i));
height[i] < height[j] ? ++i : --j;
}
return area;
}
};
Solution 3
class Solution {
public:
int maxArea(vector<int>& height) {
int area = ;
int i = , j = height.size() - ;
while (i < j) {
int h = min(height[i], height[j]);
area = max(area, (j - i) * h);
while (height[i] <= h && i < j) i++;
while (height[j] <= h && i < j) j--;
}
return area;
}
};