45. Jump Game II
Description
You are given a 0-indexed array of integers nums
of length n
. You are initially positioned at nums[0]
.
Each element nums[i]
represents the maximum length of a forward jump from index i
. In other words, if you are at nums[i]
, you can jump to any nums[i + j]
where:
0 <= j <= nums[i]
andi + j < n
Return the minimum number of jumps to reach nums[n - 1]
. The test cases are generated such that you can reach nums[n - 1]
.
Example 1:
Input: nums = [2,3,1,1,4] Output: 2 Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [2,3,0,1,4] Output: 2
Constraints:
1 <= nums.length <= 104
0 <= nums[i] <= 1000
- It's guaranteed that you can reach
nums[n - 1]
.
Solutions
Solution 1: Greedy Algorithm
We can use a variable $mx$ to record the farthest position that can be reached from the current position, a variable $last$ to record the position of the last jump, and a variable $ans$ to record the number of jumps.
Next, we traverse each position $i$ in $[0,..n - 2]$. For each position $i$, we can calculate the farthest position that can be reached from the current position through $i + nums[i]$. We use $mx$ to record this farthest position, that is, $mx = max(mx, i + nums[i])$. Then, we check whether the current position has reached the boundary of the last jump, that is, $i = last$. If it has reached, then we need to make a jump, update $last$ to $mx$, and increase the number of jumps $ans$ by $1$.
Finally, we return the number of jumps $ans$.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
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