[LeetCode]99. Container with Most Water最大容积

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.

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在第一象限存在点(1,a1),(2,a2),...,(n,an)，这些点各自与(1,0),(2,0),...,(n,0)连成一条条垂直线，任取两条(i,ai)和(j,aj)，与(i,j)组成一个容器，求能够得到的最大容积的(i,ai)和(j,aj)。

解法1：暴力破解，两层循环求出所有容器的容积，然后取最大值。Time Limit Exceeded

class Solution {
public:
int maxArea(vector<int>& height) {
int n = height.size(), maxContain = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
int minHeight = min(height[i], height[j]);
int contain = (j - i) * minHeight;
if (contain > maxContain) maxContain = contain;
}
}
return maxContain;
}
};

解法2：题目提示使用两个指针。两个指针指向数组两端，计算出容积并更新结果；然后比较左右直线的高度，将高度较小的那个移动一步；直到两个指针相遇。

class Solution {
public:
int maxArea(vector<int>& height) {
int left = 0, right = height.size() - 1, maxContain = 0;
while (left < right) {
maxContain = max(min(height[left], height[right]) * (right - left), maxContain);
if (height[left] < height[right]) ++left;
else --right;
}
return maxContain;
}
};