1. 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.
2. 思路
从最两头向中间步进。当计算[i,j]范围后,下一步推进i,j小的那一个。记录下每次的容量,选出最大的。
cont(i, j) = min{len(i), len(j)}*(j - i);
if len(i) < len(j), 则可知已i为起点的情况下,如果推进j一定比当前的小。且>j的坐标的len一定比len j小,因为每一步的推进min(len i, len j)都是非递减的。
3. 代码
耗时:26ms
class Solution {
public:
int maxArea(vector<int>& height) {
int cap = 0;
int left = 0;
int right = height.size() - 1;
while (left < right) {
int cur = (right - left) * (height[left] < height[right] ? height[left] : height[right]);
if (cur > cap) {
cap = cur;
}
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return cap;
}
};
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