45. Jump Game II

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].

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].

Solution

• 1 step에 갈 수 있는 거리를 최대화(maxRange) 하여, 마지막 index에 도착하면 step을 리턴

Code

C++

class Solution
{
public:
    int jump(vector<int> &nums)
    {
        int step = 0, start = 0, end = 0;
        while (end < nums.size() - 1)
        {
            step++;
            int maxRange = end + 1;
            for (int i = start; i <= end; i++)
            {
                if (i + nums[i] >= nums.size() - 1)
                    return step;
                maxRange = max(maxRange, i + nums[i]);
            }
            start = end + 1;
            end = maxRange;
        }
        return step;
    }
};