Pour Water
LWC 65: 755. Pour Water
Problem:
We are given an elevation map, heights[i] representing the height of the terrain at that index. The width at each index is 1. After V units of water fall at index K, how much water is at each index? Water first drops at index K and rests on top of the highest terrain or water at that index. Then, it flows according to the following rules: If the droplet would eventually fall by moving left, then move left. Otherwise, if the droplet would eventually fall by moving right, then move right. Otherwise, rise at it’s current position. Here, “eventually fall” means that the droplet will eventually be at a lower level if it moves in that direction. Also, “level” means the height of the terrain plus any water in that column. We can assume there’s infinitely high terrain on the two sides out of bounds of the array. Also, there could not be partial water being spread out evenly on more than 1 grid block - each unit of water has to be in exactly one block.
Example 1:
Input: heights = [2,1,1,2,1,2,2], V = 4, K = 3
Output: [2,2,2,3,2,2,2]
Explanation:
# #
# #
## # ###
#########
0123456 <- index
The first drop of water lands at index K = 3:
# #
# w #
## # ###
#########
0123456
When moving left or right, the water can only move to the same level or a lower level.
(By level, we mean the total height of the terrain plus any water in that column.)
Since moving left will eventually make it fall, it moves left.
(A droplet "made to fall" means go to a lower height than it was at previously.)
# #
# #
## w# ###
#########
0123456
Since moving left will not make it fall, it stays in place. The next droplet falls:
# #
# w #
## w# ###
#########
0123456
Since the new droplet moving left will eventually make it fall, it moves left.
Notice that the droplet still preferred to move left,
even though it could move right (and moving right makes it fall quicker.)
# #
# w #
## w# ###
#########
0123456
# #
# #
##ww# ###
#########
0123456
After those steps, the third droplet falls.
Since moving left would not eventually make it fall, it tries to move right.
Since moving right would eventually make it fall, it moves right.
# #
# w #
##ww# ###
#########
0123456
# #
# #
##ww#w###
#########
0123456
Finally, the fourth droplet falls.
Since moving left would not eventually make it fall, it tries to move right.
Since moving right would not eventually make it fall, it stays in place:
# #
# w #
##ww#w###
#########
0123456
The final answer is [2,2,2,3,2,2,2]:
#
#######
#######
0123456 Example 2:
Input: heights = [1,2,3,4], V = 2, K = 2
Output: [2,3,3,4]
Explanation:
The last droplet settles at index 1, since moving further left would not cause it to eventually fall to a lower height.Example 3:
Input: heights = [3,1,3], V = 5, K = 1
Output: [4,4,4]Note:
- heights will have length in [1, 100] and contain integers in [0, 99].
- V will be in range [0, 2000].
- K will be in range [0, heights.length - 1].
思路：
模拟，每次drop一滴水，接着从左边开始找坑，坑的定义如下，存在j，使得a[j] < a[j + 1] && a[j - 1] > a[j] ， 找到j则填坑，找不到就从右边开始找坑，找到填坑，找不到填位置K。
Java版本：
public int[] pourWater(int[] a, int V, int K) {
int n = a.length;
for (int i = 0; i < V; ++i) {
boolean ok = false;
int index = -1;
for (int j = K - 1; j >= 0; --j) {
if (a[j] > a[j + 1]) break;
if (a[j] < a[j + 1]) {
ok = true;
index = j;
}
}
if (ok) {
a[index] ++;
continue;
}
for (int j = K + 1; j < n; ++j) {
if (a[j] > a[j - 1]) break;
if (a[j] < a[j - 1]) {
ok = true;
index = j;
}
}
if (ok) a[index] ++; else a[K] ++;
}
return a;
}Python版本：
def pourWater(self, a, V, K):
"""
:type heights: List[int]
:type V: int
:type K: int
:rtype: List[int]
"""
n = len(a)
for i in range(V):
ok = False
index = -1
for j in range(K - 1, -1, -1):
if a[j] > a[j + 1]: break
if a[j] < a[j + 1]:
ok = True
index = j
if ok:
a[index] += 1
continue
for j in range(K + 1, n):
if a[j] > a[j - 1]: break
if a[j] < a[j - 1]:
ok = True
index = j
if ok:
a[index] += 1
else:
a[K] += 1
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