Leetcode: 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!

这道题做的比较艰辛，一开始自己想的是一个用stack的解法，感觉过于繁琐（出栈，入栈，计算容积），但未尝不是一个好的尝试，这个方法还是有点小问题，过后会好好想清楚。看了网上提示完成了最终的方法，这个方法两次遍历数组，第一次遍历找每个元素右边最大的元素，第二次遍历寻找每个元素左边最大的元素，同时计算该index可以存放的水容积： min{lefthighest, righthighest}-A[i]

第二遍方法：

 1 public class Solution {

 2     public int trap(int[] A) {

 3         int[] left = new int[A.length];

 4         int[] right = new int[A.length];

 5         int sum = 0;

 6         for (int i=0; i<A.length; i++) {

 7             if (i == 0) left[i] = 0;

 8             else {

 9                 left[i] = Math.max(left[i-1], A[i-1]);

10             }

11         }

12         for (int i=A.length-1; i>=0; i--) {

13             if (i == A.length - 1) right[i] = 0;

14             else {

15                 right[i] = Math.max(right[i+1], A[i+1]);

16             }

17             if (Math.min(left[i], right[i]) > A[i]) {

18                 sum += Math.min(left[i], right[i]) - A[i];

19             }

20         }

21         return sum;

22     }

23 }

之前用stack尝试解的方法，较复杂，但是是个不错的思考切入点：

 1 public class Solution {

 2     public int trap(int[] A) {

 3         Stack<Integer> s = new Stack<Integer>();

 4         int i = 0, sum = 0, lastpop = 0;

 5         while (i < A.length){

 6             if (s.empty()) {

 7                 s.push(i);

 8                 i++;

 9                 continue;

10             }

11             if (A[s.peek()] < A[i]){ //can hold water if stack contains other elements

12                 while (A[s.peek()] < A[i]){

13                     lastpop = A[s.pop()];

14                     if (s.empty()){

15                         s.push(i);

16                         i++;

17                         continue;

18                     }

19                     sum += (i - s.peek() -1) * (A[s.peek()] - lastpop);

20                 }

21                 if (A[s.peek()] > A[i]){

22                     sum += (i - s.peek() -1) * (A[i] - lastpop);

23                     s.push(i);

24                     i++;

25                     continue;

26                 }

27                 if (A[s.peek()] == A[i]){

28                     sum += (i - s.peek() -1) * (A[i] - lastpop);

29                     s.pop();

30                     i++;

31                     continue;

32                 }

33             }

34             else if (s.peek() == A[i]){

35                 i++;

36                 continue;

37             }

38             else if (s.peek() > A[i]){

39                 s.push(i);

40                 i++;

41                 continue;

42             }

43         }

44         return sum;

45     }

46 }