You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return the maximum amount of water a container can store.
Notice that you may not slant the container.
Example 1:
Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Explanation: 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 2:
Input: height = [1,1] Output: 1
Constraints:
n == height.length2 <= n <= 1050 <= height[i] <= 104class Solution {
public:
int maxArea(vector<int>& ht) {
int maxm=0, lt=0, rt=ht.size()-1;
while(lt<rt){
maxm = max(maxm, (rt-lt)*min(ht[lt], ht[rt]));
ht[lt]<ht[rt]?lt++:rt--;
}
return maxm;
}
};class Solution {
public:
int maxArea(vector<int>& ht) {
int maxm=0, lt=0, rt=ht.size()-1;
while(lt<rt){
maxm = max(maxm, (rt-lt)*min(ht[lt], ht[rt]));
if(ht[lt]<ht[rt])
lt++;
else rt--;
}
return maxm;
}
};class Solution {
public int maxArea(int[] height) {
int lt = 0, rt = height.length - 1, ans = 0;
while (lt < rt) {
if (height[lt] < height[rt])
ans = Math.max(ans, (rt - lt) * height[lt++]);
else
ans = Math.max(ans, (rt - lt) * height[rt--]);
}
return ans;
}
}var maxArea = function(ht) {
var maxm=0, lt=0, rt=ht.length-1;
while(lt<rt){
maxm = Math.max(maxm,(rt-lt)*Math.min(ht[lt], ht[rt]));
if(ht[lt]<ht[rt])
lt++;
else rt--;
}
return maxm;
};