LeetCode 042 Trapping Rain Water

题目

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

1  难点是找到雨水积累的规律。

2 单个点X是否积水，要看左边的最高高度和右边的最高高度是否都比X的高度高；如果都高，那么肯定积水

3 单个点X积多少水？选择左边最高高度和右边最高高度当中最小的一个Y，Y-X即为这点积水。

4 由此可以得到思路。做两个数组来记录某点的左右最高高度即可。

代码

public class Solution {
public int trap(int[] A) {
if(A.length<=2){
return 0;
}
int n = A.length;
int[] left = new int
;
int[] right = new int
;
int max =0;
for(int i=1;i<n;i++){
if(A[i-1]>max){
max = A[i-1];
}
left[i]=max;
}
max =0;
for(int i=n-2;i>=0;i--){
if(A[i+1]>max){
max = A[i+1];
}
right[i]=max;
}
int sum =0;
for(int i=0;i<n;i++){
if(A[i]<Math.min(left[i],right[i])){
sum+=Math.min(left[i],right[i])-A[i];
}
}
return sum;
}
}