Description
Tag：Dynamic Programming
Difficulty：Hard
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 < 1,100.
- Each stone's position will be a non-negative integer < 231.
- 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.
Solution
一開始以爲這是一個 backtracking 的問題，確實能解決，但是時間複雜度太高，在 999999 那個case 裏報了 TLE，試了許久，包括 worst case 過濾(某個網友的答案就是這種), 二分查找過濾，最後看到標籤才意識到需要使用DP，最終accepted。
這個網友的答案,記錄每次蛙跳後到達石頭可以再次跳躍的步數，也是非常棒的一個思路
自己的代碼：
var stepStoneMap map[string]int
func canCross(stones []int) bool {
stepStoneMap = make(map[string]int)
return backtrack(stones, []int{1})
}
func backtrack(stones []int, units []int) bool {
if len(stones) <= 1 {
return true
}
//all steps can take
for _,unit := range units {
var temp int = unit
if unit >= len(stones) {
temp = len(stones) - 1
}
//jump
for jump:=temp;jump>=1;jump--{
key := strconv.Itoa(jump) + "," + strconv.Itoa(stones[jump])
// 剪枝
if _,ok:=stepStoneMap[key];ok{
continue
}
if stones[jump] - stones[0] == unit {
stepStoneMap[key] = 1
if backtrack(stones[jump:], jumpUnits(unit)) {
return true
}
}
}
}
return false
}
func jumpUnits(k int) []int{
if k == 1 {
return []int{1, 2}
}
return []int{k+1, k, k-1}
}