55.Jump Game:修订间差异

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(未显示同一用户的2个中间版本)
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|category=Dynamic Programming
|category=Dynamic Programming
|collection=Top 150
|collection=Top 150
|title=Triangle
|title=Jump Game
|summary=You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.
|summary=You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.


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         }
         }
         return dp[nums.length-1];
         return dp[nums.length-1];
    }
}
</syntaxhighlight>
== Greedy ==
{{Submission|runtime=2ms|memory=44.02MB|rp=80.65|mp=69.15}}
<syntaxhighlight lang="bash">
class Solution {
    public boolean canJump(int[] nums) {
        // keep track of the current max reachable position of each step
        int reachable = 0;
        for(int i = 0; i < nums.length; i++) {
            if(i > reachable) return false;
            reachable = Math.max(reachable, i + nums[i]);
        }
        return true;
     }
     }
}
}
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[[Category:Algorithm]]
[[Category:Algorithm]]
[[Category:Dynamic Programming]]
[[Category:Dynamic Programming]]
[[Category:Greedy]]
[[Category:LeetCode]]
[[Category:LeetCode]]

2023年11月22日 (三) 15:16的最新版本

Description

#55 Jump Game Medium
Dynamic Programming Top 150
You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position. Return true if you can reach the last index, or false otherwise.

Example 1:

Input: nums = [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: nums = [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

1D DP

Runtime 418ms
Memory 43.4MB
class Solution {
    public boolean canJump(int[] nums) {
        /*
            [2,3,1,1,4]
             1 1 2 2 2
         */
        int []dp = new int[nums.length];
        dp[0] = 1;
        for(int i = 0; i < nums.length; i++) {
            for(int j = 1; j <= nums[i]; j++){
                if(i+j >= dp.length) break;
                dp[i+j]++;
            }
                
        }
        for(int i = 0; i < dp.length; i++) {
            if(dp[i] == 0) return false;
        }
            
        return true;
    }
}

1D DP 2

Runtime 76ms
Memory 43.85MB
class Solution {
    public boolean canJump(int[] nums) {
        boolean []dp = new boolean[nums.length];
        dp[0] = true;
        for(int i = 0; i <nums.length; i++) {
            if(!dp[i]) return false;
            for(int j = 1; j <= nums[i]; j++) {
                int next = i + j;
                if(next >= dp.length) break;
                if(next == nums.length -1) return true; // reach end already
                dp[next] = true;
            }
        }
        return dp[nums.length-1];
    }
}

Greedy

Runtime 2ms
Memory 44.02MB
class Solution {
    public boolean canJump(int[] nums) {
        // keep track of the current max reachable position of each step
        int reachable = 0;
        for(int i = 0; i < nums.length; i++) {
            if(i > reachable) return false;
            reachable = Math.max(reachable, i + nums[i]);
        }
        return true;
    }
}