7_leetcode_container with most water
2014-06-04 08:18
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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.
//1:考虑特殊情况,2:前后两个指针,算出两个指针之间的水量; 3:当前指向的水少的指针向另外一个指针靠近
int maxArea(vector<int> &height)
{
if(height.size() <= 1)
return 0;
int water = 0;
int left = 0;
int right = (int)height.size() - 1;
while(left < right)
{
int distance = right - left;
int temp = (height[left] > height[right] ? height[right] : height[left]);
if(temp * distance > water)
{
water = temp * distance;
}
if(height[left] < height[right])
{
left++;
}
else
{
right--;
}
}
return water;
}
..., 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.
//1:考虑特殊情况,2:前后两个指针,算出两个指针之间的水量; 3:当前指向的水少的指针向另外一个指针靠近
int maxArea(vector<int> &height)
{
if(height.size() <= 1)
return 0;
int water = 0;
int left = 0;
int right = (int)height.size() - 1;
while(left < right)
{
int distance = right - left;
int temp = (height[left] > height[right] ? height[right] : height[left]);
if(temp * distance > water)
{
water = temp * distance;
}
if(height[left] < height[right])
{
left++;
}
else
{
right--;
}
}
return water;
}
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