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LeetCode--Container With Most Water

2017-07-16 10:06 429 查看
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.

思路:双指针,贪心策略。先确定扫描长度是最长的,也就是从两端扫描起,找到两端较短的木板后,往中间靠拢,使短板变长,再计算更新最大面积,当双指针在中间汇集时结束。

木桶效应:装水的面积由短板决定。

class Solution {
public:
int maxArea(vector<int>& height) {
int water = 0;
int i = 0, j = height.size() - 1;
while (i < j) {
int h = min(height[i], height[j]);
water = max(water, (j - i) * h);
while (height[i] <= h && i < j) i++;
while (height[j] <= h && i < j) j--;
}
return water;
}
};
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