403. Frog Jump
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
一开始以为这是一个 backtracking 的问题，确实能解决，但是时间复杂度太高，在 999999 那个case 里报了 TLE，试了许久，包括 worst case 过滤(某个网友的答案就是这种), 二分查找过滤，最后看到标签才意识到需要使用DP，最终accepted。
这个网友的答案,记录每次蛙跳后到达石头可以再次跳跃的步数，也是非常棒的一个思路
自己的代码：
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}
}