code/leetcode/v1/jumpgame

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Jump Game

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

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

Determine if you are able to reach the last index.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
jump length is 0, which makes it impossible to reach the last index.

Solution 1

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class Solution {
    public boolean canJump(int[] nums) {
        int[] tag = new int[nums.length];
        tag[nums.length - 1] = 1;
        for (int i = nums.length - 2; i >= 0; i--) {
            for (int j = i + 1; j <= i + nums[i] && j < nums.length; j++) {
                if (tag[j] == 1) {
                    tag[i] = 1;
                    break;
                }
            }
        }

        return tag[0] == 1;
    }
}
/*
执行用时 :580 ms, 在所有 Java 提交中击败了11.08%的用户
内存消耗 :40.6 MB, 在所有 Java 提交中击败了31.49%的用户
*/

Solution 2

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class Solution {
    public boolean canJump(int[] nums) {
        int res = nums.length - 1;
        
        for (int i = nums.length - 1; i >= 0; i--) {
            if (i + nums[i] >= res) {
                res = i;
            }
        }
        
        return res == 0;
    }
}
/*
执行用时 :1 ms, 在所有 Java 提交中击败了99.95%的用户
内存消耗 :40.9 MB, 在所有 Java 提交中击败了24.65%的用户
*/

We can use the position of that we can reach the target to record current result.
So we just need to calculate whether current position plus current value >= last which can reach the target result, so we can tag current position.

Reference
https://leetcode-cn.com/problems/jump-game