描述
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複製程式碼思路
生成一個二維陣列，計算每兩個擋板之間的存水量，時間複雜度均為O(n^2)。
class Solution {
public int maxArea(int[] height) {
int out = 0;
for(int i = 0; i<height.length; i++){
for(int j = 0; j<=i; j++){
out = Math.max(out,((i - j) * Math.min(height[i], height[j])));
}
}
return out;
}
}複製程式碼可以看出時間消耗確實很高。
優化
評論區發現一種思路，當有一邊的擋板高度大於另一邊時，將高的那邊的index向中間收縮不會得到更大的值，因為寬和高一定都會縮短。
The max area is calculated by the following formula:
S = (j - i) * min(ai, aj)
We should choose (i, j) so that S is max. Note that
i, jgo through the range (1, n) and j > i. That's it.
The simple way is to take all possibilities of (i, j) and compare all obtained S. The time complexity is
n * (n-1) / 2
What we gonna do is to choose all possibilities of (i, j) in a wise way. I noticed that many submitted solutions here can't explain why when :
ai < ajwe will check the next(i+1, j)(or move i to the right)ai >= ajwe will check the next(i, j-1)(or move j to the left)Here is the explaination for that:
- When
ai < aj, we don't need to calculate all(i, j-1),(i, j-2), .... Why? because these max areas are smaller than our S at(i, j)Proof: Assume at
(i, j-1)we haveS'= (j-1-i) * min(ai, aj-1)
S'< (j-1-i) * ai < (j-i) * ai = S, and whenS'<S, we don't need to calculate
Similar at(i, j-2),(i, j-3), etc.
So, that's why when
ai < aj, we should check the next at(i+1, j)(or move i to the right)
- When
ai >= aj, the same thing, all(i+1, j),(i+2, j), .... are not needed to calculate.We should check the next at
(i, j-1)(or move j to the left)
下面是實現：
class Solution {
public int maxArea(int[] height) {
int i = 0, j = height.length -1, s = 0;
while(i != j){
s = Math.max(s, ((j - i) * Math.min(height[i], height[j])));
if(height[i] < height[j]){
i++;
}else{
j--;
}
}
return s;
}
}複製程式碼避免迴圈還是能夠節約很多時間的。