【leetcode】Container With Most Water
2014-08-26 23:39
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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.
解析:给一组数组作为水池的高度,以其中两个高度作为水池的壁,求最大的水池容量。假设左边的壁为left,右边的壁为right,则水池容量:
capacity= (right-left)*min(height(right),height(right))
解决方法是定义两个指针left和right从数组两头开始遍历:如果left的高度低于right,则将left指针右移找到比left高的重新判定capacity;反之,如果right的高度低于left,则将right的指针左移找到比right高的重新判定capacity。最后left==right结束while循环。
C++ AC代码:
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.
解析:给一组数组作为水池的高度,以其中两个高度作为水池的壁,求最大的水池容量。假设左边的壁为left,右边的壁为right,则水池容量:
capacity= (right-left)*min(height(right),height(right))
解决方法是定义两个指针left和right从数组两头开始遍历:如果left的高度低于right,则将left指针右移找到比left高的重新判定capacity;反之,如果right的高度低于left,则将right的指针左移找到比right高的重新判定capacity。最后left==right结束while循环。
C++ AC代码:
class Solution { public: int maxArea(vector<int> &height) { int capacity = 0; int left = 0,right = height.size()-1; int k; while(left<right){ capacity = max(capacity,(right-left)*min(height[left],height[right])); if(height[left]<height[right]){ k = left; while(k<right && height[k] <= height[left]){ k++; } left = k; }else{ k = right; while(k>left && height[k] <= height[right]){ k--; } right = k; } } return capacity; } };
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