#algo/greedy ## Notes - 贪心算法，贪心的原因是对于 `nums[i]` 所对应的那个数字，跳到那里之后你可以接下来跳的是一个 range。所以，并不会有什么额外的限制，肯定是越跳越多越好。 ## Problem ### Problem 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]` and - `i + 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]`. ### Examples #### 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 <= 10^4` - `0 <= nums[i] <= 1000` - It's guaranteed that you can reach `nums[n - 1]`.