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leetcode/algorithms/cpp/frogJump/FrogJump.cpp at master · fallleaves/leetcode · GitHub
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//
Source : https://leetcode.com/problems/frog-jump/
//
Author : Hao Chen
//
Date : 2016-11-12
/*
**************************************************************************************
*
* A frog is crossing a river. The river is divided into x units and at each unit there
* may or may not exist a stone. The frog can jump on a stone, but it must not jump
* into the water.
*
* Given a list of stones' positions (in units) in sorted ascending order, determine if
* the frog is able to cross the river by landing on the last stone. Initially, the
* frog is on the first stone and assume the first jump must be 1 unit.
*
* If the frog's last jump was k units, then its next jump must be either k - 1, k, or
* k + 1 units. Note that the frog can only jump in the forward direction.
*
* Note:
*
* The number of stones is ≥ 2 and is
* Each stone's position will be a non-negative integer 31.
* The first stone's position is always 0.
*
* Example 1:
*
* [0,1,3,5,6,8,12,17]
*
* There are a total of 8 stones.
* The first stone at the 0th unit, second stone at the 1st unit,
* third stone at the 3rd unit, and so on...
* The last stone at the 17th unit.
*
* Return true. The frog can jump to the last stone by jumping
* 1 unit to the 2nd stone, then 2 units to the 3rd stone, then
* 2 units to the 4th stone, then 3 units to the 6th stone,
* 4 units to the 7th stone, and 5 units to the 8th stone.
*
* Example 2:
*
* [0,1,2,3,4,8,9,11]
*
* Return false. There is no way to jump to the last stone as
* the gap between the 5th and 6th stone is too large.
**************************************************************************************
*/
class
Solution
{
public:
bool
canCross_recursion
(vector<
int
>& stones,
int
curr,
int
last_jump) {
for
(
int
i=curr+
1
; i<stones.
size
(); i++){
int
next_jump = stones[i] - stones[curr];
//
the minimal jump is far exceed the current node, go to check next node.
if
(next_jump < last_jump -
1
)
continue
;
//
cannot reach this one, then simple reture false;
if
(next_jump > last_jump +
1
)
return
false
;
if
(i == stones.
size
() -
1
||
canCross_recursion
(stones, i, next_jump))
return
true
;
}
return
false
;
}
bool
canCross_recursion_with_cache
(vector<
int
>& stones,
int
curr,
int
last_jump,
unordered_map<
int
, unordered_map<
int
,
bool
>>& cache)
{
//
check the cache is hitted ?
if
(cache.
find
(curr) != cache.
end
() && cache[curr].
find
(last_jump)!=cache[curr].
end
()) {
return
cache[curr][last_jump];
}
for
(
int
i=curr+
1
; i<stones.
size
(); i++){
int
next_jump = stones[i] - stones[curr];
if
(next_jump < last_jump -
1
)
continue
;
if
(next_jump > last_jump +
1
)
break
;
if
(i == stones.
size
() -
1
||
canCross_recursion_with_cache
(stones, i, next_jump, cache)) {
cache[curr][last_jump] =
true
;
return
true
;
}
}
cache[curr][last_jump] =
false
;
return
false
;
}
bool
canCross_non_recursion
(vector<
int
>& stones) {
//
the `jumps` map store the all possible `last jumps`
unordered_map<
int
, unordered_set<
int
>> jumps = {{
0
, {
0
}}};
for
(
int
i=
0
; i<stones.
size
(); i++) {
if
(jumps.
find
(i) == jumps.
end
()){
continue
;
}
//
for each possible last jump which reach the current node.
for
(
int
last_jump : jumps[i]) {
//
find the next nodes can be reached.
for
(
int
j=i+
1
; j < stones.
size
(); j++) {
//
ingore the rest node which cannot be reached
if
(stones[i] + last_jump +
1
< stones[j])
break
;
//
evaluated three possbile jumps for next node
for
(
int
next_jump = last_jump -
1
; next_jump <= last_jump +
1
; next_jump++) {
if
( stones[i] + next_jump == stones[j] ) {
jumps[j].
insert
(next_jump);
}
}
}
}
}
return
jumps.
find
(stones.
size
()-
1
)!=jumps.
end
();
}
bool
canCross
(vector<
int
>& stones) {
//
Burst Force solution -- accepted ~500ms
return
canCross_non_recursion
(stones);
//
DFS with cache solution - accepted ~160ms
unordered_map<
int
, unordered_map<
int
,
bool
>> cache;
return
canCross_recursion_with_cache
(stones,
0
,
0
, cache);
//
Time Limit Error
return
canCross_recursion
(stones,
0
,
0
);
}
};
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