Container With Most Water
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.
分析参考这里，作者讲解的很详细
class Solution {
public:
int maxArea(vector<int> &height) {
int left = 0,right = height.size()-1,area = 0;
while(left < right)
{
area = max(area,min(height[left],height[right])*(right-left));//当前位置的面积
if(height[left] < height[right])
{
int k = left;
while(k < right && height[k] <= height[left])++k;//找下一个位置
left = k;
}
else
{
int k = right;
while(k > left && height[k] <= height[right])--k;//找下一个位置
right = k;
}
}
return area;
}
};
Trapping Rain Water
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
思路见这里，作者讲解的很详细
class Solution {
public:
int trap(int A[], int n)
{
vector<int> leftMostHeight(n,0),rightMostHeight(n,0);
int mostHeiht = 0,i,area = 0;
for(i = 0;i < n;++i)
{
leftMostHeight[i] = mostHeiht;//左边的最大高度
mostHeiht = max(mostHeiht,A[i]);
}
mostHeiht = 0;
for(i = n-1;i >= 0;--i)
{
rightMostHeight[i] = mostHeiht;//右边的最大高度
mostHeiht = max(mostHeiht,A[i]);
}
for(i = 0;i < n;++i)
{
int a = min(leftMostHeight[i],rightMostHeight[i]) - A[i];//当前高度的蓄水量
if(a > 0)area += a;
}
return area;
}
};
另外还有两个相似的题目，都是和直方图相关的，具体看这里