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]
.
题目含义:
var jump = function (nums) {
let ans = 0, jp = 0, far = 0
for (let i = 0; i < nums.length - 1; i++) {
jp = Math.max(jp, nums[i] + i)
if (i === far) {
far = jp
ans++
}
}
return ans
};