Trapping Rain Water (Bar Height) -- LeetCode
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。
1 class Solution {
2 public:
3 int trap(vector<int>& height) {
4 if (height.size() == 0) return 0;
5 int left = 0, right = height.size() - 1, res = 0;
6 int maxleft = height[left], maxright = height[right];
7 while (left < right)
8 {
9 if (height[left] <= height[right])
10 {
11 if (height[left] > maxleft) maxleft = height[left];
12 else res += maxleft - height[left];
13 left++;
14 }
15 else
16 {
17 if (height[right] > maxright) maxright = height[right];
18 else res += maxright - height[right];
19 right--;
20 }
21 }
22 return res;
23 }
24 };
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。
1 class Solution {
2 public:
3 int trap(vector<int>& height) {
4 if (height.size() == 0) return 0;
5 int left = 0, right = height.size() - 1, res = 0;
6 int maxleft = height[left], maxright = height[right];
7 while (left < right)
8 {
9 if (height[left] <= height[right])
10 {
11 if (height[left] > maxleft) maxleft = height[left];
12 else if (maxleft - height[left] > 0)
13 res = max(res, maxleft);
14 left++;
15 }
16 else
17 {
18 if (height[right] > maxright) maxright = height[right];
19 else if (maxright - height[right] > 0)
20 res = max(res, maxright);
21 right--;
22 }
23 }
24 return res;
25 }
26 };