Update 0045.跳跃游戏II.md

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jianghongcheng
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@ -205,66 +205,81 @@ class Solution {
```
### Python
贪心(版本一)
```python
class Solution:
def jump(self, nums: List[int]) -> int:
if len(nums) == 1: return 0
ans = 0
curDistance = 0
nextDistance = 0
for i in range(len(nums)):
nextDistance = max(i + nums[i], nextDistance)
if i == curDistance:
if curDistance != len(nums) - 1:
ans += 1
curDistance = nextDistance
if nextDistance >= len(nums) - 1: break
return ans
```
```python
# 贪心版本二
class Solution:
def jump(self, nums: List[int]) -> int:
def jump(self, nums):
if len(nums) == 1:
return 0
curDistance, nextDistance = 0, 0
step = 0
for i in range(len(nums)-1):
nextDistance = max(nextDistance, nums[i]+i)
if i == curDistance:
curDistance = nextDistance
step += 1
return step
```
```python
# 贪心版本三 - 类似55-跳跃游戏’写法
class Solution:
def jump(self, nums) -> int:
if len(nums)==1: return 0
i = 0
count = 0
cover = 0
while i<=cover:
for i in range(i,cover+1):
cover = max(nums[i]+i,cover)
if cover>=len(nums)-1: return count+1
count+=1
cur_distance = 0 # 当前覆盖最远距离下标
ans = 0 # 记录走的最大步数
next_distance = 0 # 下一步覆盖最远距离下标
for i in range(len(nums)):
next_distance = max(nums[i] + i, next_distance) # 更新下一步覆盖最远距离下标
if i == cur_distance: # 遇到当前覆盖最远距离下标
ans += 1 # 需要走下一步
cur_distance = next_distance # 更新当前覆盖最远距离下标(相当于加油了)
if next_distance >= len(nums) - 1: # 当前覆盖最远距离达到数组末尾不用再做ans++操作,直接结束
break
return ans
```
贪心(版本二)
```python
# 动态规划做法
class Solution:
def jump(self, nums):
cur_distance = 0 # 当前覆盖的最远距离下标
ans = 0 # 记录走的最大步数
next_distance = 0 # 下一步覆盖的最远距离下标
for i in range(len(nums) - 1): # 注意这里是小于len(nums) - 1这是关键所在
next_distance = max(nums[i] + i, next_distance) # 更新下一步覆盖的最远距离下标
if i == cur_distance: # 遇到当前覆盖的最远距离下标
cur_distance = next_distance # 更新当前覆盖的最远距离下标
ans += 1
return ans
```
贪心(版本三) 类似55-跳跃游戏’写法
```python
class Solution:
def jump(self, nums) -> int:
if len(nums)==1: # 如果数组只有一个元素不需要跳跃步数为0
return 0
i = 0 # 当前位置
count = 0 # 步数计数器
cover = 0 # 当前能够覆盖的最远距离
while i <= cover: # 当前位置小于等于当前能够覆盖的最远距离时循环
for i in range(i, cover+1): # 遍历从当前位置到当前能够覆盖的最远距离之间的所有位置
cover = max(nums[i]+i, cover) # 更新当前能够覆盖的最远距离
if cover >= len(nums)-1: # 如果当前能够覆盖的最远距离达到或超过数组的最后一个位置,直接返回步数+1
return count+1
count += 1 # 每一轮遍历结束后,步数+1
```
动态规划
```python
class Solution:
def jump(self, nums: List[int]) -> int:
result = [10**4+1]*len(nums)
result[0]=0
for i in range(len(nums)):
for j in range(nums[i]+1):
if i+j<len(nums): result[i+j]=min(result[i+j],result[i]+1)
#print(result) #打印数组
return result[-1]
result = [10**4+1] * len(nums) # 初始化结果数组,初始值为一个较大的数
result[0] = 0 # 起始位置的步数为0
for i in range(len(nums)): # 遍历数组
for j in range(nums[i] + 1): # 在当前位置能够跳跃的范围内遍历
if i + j < len(nums): # 确保下一跳的位置不超过数组范围
result[i + j] = min(result[i + j], result[i] + 1) # 更新到达下一跳位置的最小步数
return result[-1] # 返回到达最后一个位置的最小步数
```