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:使用两个指针left和right,分别从两端向中间靠拢。同时,记录left和right曾经到过的最高值。如果当前值并没有超过最高值,则将最高值与当前的差值作为水的量,并将指针加一(right指针是减一)。为了让这个方法有效(无效的情况是,指针后续遇见的bar都不会再高于最高值了,那么这些水根本存不起来),因此我们每次只移动left和right中矮的那一个(如果两者相等,则移动left)。这样,一定能保证最后会碰到比最高值要高的bar(至少是等高),因为我们移动的是矮的那个指针,往前移动肯定最后会遇见另一个高的指针。

 class Solution {
public:
int trap(vector<int>& height) {
if (height.size() == ) return ;
int left = , right = height.size() - , res = ;
int maxleft = height[left], maxright = height[right];
while (left < right)
{
if (height[left] <= height[right])
{
if (height[left] > maxleft) maxleft = height[left];
else res += maxleft - height[left];
left++;
}
else
{
if (height[right] > maxright) maxright = height[right];
else res += maxright - height[right];
right--;
}
}
return res;
}
};

bar height问题:这是Amazon面试中的一道问题。求最高的液面高度。这里只需要把上方的代码修改一下就可以用。就是在res值变更时,若max与当前高度差值非零时,记录下max值,最后最高的max值就是结果。

 class Solution {
public:
int trap(vector<int>& height) {
if (height.size() == ) return ;
int left = , right = height.size() - , res = ;
int maxleft = height[left], maxright = height[right];
while (left < right)
{
if (height[left] <= height[right])
{
if (height[left] > maxleft) maxleft = height[left];
else if (maxleft - height[left] > )
res = max(res, maxleft);
left++;
}
else
{
if (height[right] > maxright) maxright = height[right];
else if (maxright - height[right] > )
res = max(res, maxright);
right--;
}
}
return res;
}
};

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