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]
.
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.
Input: nums = [2,3,0,1,4] Output: 2
1 <= nums.length <= 104
0 <= nums[i] <= 1000
- It's guaranteed that you can reach
nums[n - 1]
.
impl Solution {
pub fn jump(nums: Vec<i32>) -> i32 {
let mut i = 0;
let mut j = 0;
let mut ret = 0;
for k in 0..nums.len() {
if k > i {
i = j;
ret += 1;
}
j = j.max(k + nums[k] as usize);
}
ret
}
}