README_EN.md

45. Jump Game II

中文文档

Description

Given an array of non-negative integers nums, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

You can assume that you can always reach the last index.

 

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 <= 1000
• 0 <= nums[i] <= 105

Solutions

Python3

class Solution:
    def jump(self, nums: List[int]) -> int:
        end = mx = steps = 0
        for i, num in enumerate(nums[:-1]):
            mx = max(mx, i + num)
            if i == end:
                end = mx
                steps += 1
        return steps

Java

class Solution {
    public int jump(int[] nums) {
        int end = 0;
        int mx = 0;
        int steps = 0;
        for (int i = 0; i < nums.length - 1; ++i) {
            mx = Math.max(mx, i + nums[i]);
            if (i == end) {
                end = mx;
                ++steps;
            }
        }
        return steps;
    }
}

C++

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

Go

func jump(nums []int) int {
	mx, steps, end := 0, 0, 0
	for i := 0; i < len(nums)-1; i++ {
		mx = max(mx, i+nums[i])
		if i == end {
			end = mx
			steps++
		}
	}
	return steps
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

C#

public class Solution {
    public int Jump(int[] nums) {
        int end = 0;
        int mx = 0;
        int steps = 0;
        for (int i = 0; i < nums.Length - 1; ++i)
        {
            mx = Math.Max(mx, i + nums[i]);
            if (i == end)
            {
                end = mx;
                ++steps;
            }
        }
        return steps;
    }
}

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