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!

 

思想：积水高度右左右最小高度以及自己高度决定，可以得到如下结论  sum +=- min(lheight[i],rheight[i]) - A[i] > 0

java代码：

1. public int trap(int[] A) {
2. int size = A.length;
3. int[] lheight = new int[size];
4. int[] rheight = new int[size];
5. int max=0;
6. for(int i=0;i<size;i++) {
7. //if(i==0) height[i]=A[i];
8. lheight[i]=max;
9. if(A[i]>max) {
10. max=A[i];
11. lheight[i]=max;
12. }
13. }
14. max=0;
15. for(int i=size-1;i>=0;i--) {
16. rheight[i]=max;
17. if(A[i]>max) {
18. max=A[i];
19. rheight[i]=max;
20. }
21. }
22. int sum=0;
23. for(int i=1;i<size-1;i++) {
24. int minmax = Math.min(lheight[i-1],rheight[i+1]);
25. if(minmax>A[i]) sum+=(minmax-A[i]);
26. }
27. return sum;
28. }

C++代码：

1. int trap(int A[], int n) {
2. if(n==0) return 0;
3. int water=0;
4. int curHeight=0;
5. int i=0;
6. int mid=0;
7. for(i=0;i<n;i++) {
8. if(A[i]>A[mid]) {
9. mid=i;
10. }
11. }
12. for(i=0;i<mid;i++) {
13. if(A[i]<curHeight) {
14. water+=(curHeight-A[i]);
15. } else curHeight=A[i];
16. }
17. curHeight=0;
18. for(i=n-1;i>mid;i--) {
19. if(A[i]<curHeight) water+=(curHeight-A[i]);
20. else curHeight=A[i];
21. }
22. return water;
23. }