https://oj.leetcode.com/problems/trapping-rain-water/
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!
SOLUTION 1:
从左到右扫描，计算到从左边到curr的最高的bar,从右到左扫描，计算到从右边到curr的最高的bar。
再扫描一次，把这两者的低者作为{桶}的高度，如果这个桶高于A[i]的bar,那么A[i]这个bar上头可以存储height - A[i]这么多水。把这所有的水加起来即可。
public class Solution {
public int trap(int[] A) {
if (A == null) {
return ;
}
int max = ;
int len = A.length;
int[] left = new int[len];
int[] right = new int[len];
// count the highest bar from the left to the current.
for (int i = ; i < len; i++) {
left[i] = i == ? A[i]: Math.max(left[i - ], A[i]);
}
// count the highest bar from right to current.
for (int i = len - ; i >= ; i--) {
right[i] = i == len - ? A[i]: Math.max(right[i + ], A[i]);
}
// count the largest water which can contain.
for (int i = ; i < len; i++) {
int height = Math.min(right[i], left[i]);
if (height > A[i]) {
max += height - A[i];
}
}
return max;
}
}
2015.1.14 redo:
合并2个for 循环，简化计算。
public class Solution {
public int trap(int[] A) {
// 2:37
if (A == null) {
return ;
}
int len = A.length;
int[] l = new int[len];
int[] r = new int[len];
for (int i = ; i < len; i++) {
if (i == ) {
l[i] = A[i];
} else {
l[i] = Math.max(l[i - ], A[i]);
}
}
int water = ;
for (int i = len - ; i >= ; i--) {
if (i == len - ) {
r[i] = A[i];
} else {
// but: use Math, not max
r[i] = Math.max(r[i + ], A[i]);
}
water += Math.min(l[i], r[i]) - A[i];
}
return water;
}
}
CODE:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/Trap.java