Container With Most Water
2015-08-01 17:34
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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.
典型的双指针应用问题,令low = 0, high = len - 1,然后向中间靠拢,向中间靠拢意味着宽在减少,那为了使面积更加得大,唯一的可能是找到更高的高度。class Solution {
public:
//双指针
int maxArea(vector<int>& height) {
int len = height.size();
if(len < 2)
return 0;
int low = 0, high = len - 1;
int max = -1;
while(low < high)
{
int tmpArea = (high - low) * min(height[low], height[high]);
max = (max < tmpArea) ? tmpArea : max;
if(height[low] <= height[high])
++low;
else
--high;
}
return max;
}
};
..., 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.
典型的双指针应用问题,令low = 0, high = len - 1,然后向中间靠拢,向中间靠拢意味着宽在减少,那为了使面积更加得大,唯一的可能是找到更高的高度。class Solution {
public:
//双指针
int maxArea(vector<int>& height) {
int len = height.size();
if(len < 2)
return 0;
int low = 0, high = len - 1;
int max = -1;
while(low < high)
{
int tmpArea = (high - low) * min(height[low], height[high]);
max = (max < tmpArea) ? tmpArea : max;
if(height[low] <= height[high])
++low;
else
--high;
}
return max;
}
};
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