55.Jump Game:修订间差异
(未显示同一用户的4个中间版本) | |||
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|category=Dynamic Programming | |category=Dynamic Programming | ||
|collection=Top 150 | |collection=Top 150 | ||
|title= | |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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==1D DP 2== | ==1D DP 2== | ||
{{Submission|runtime= | {{Submission|runtime=76ms|memory=43.85MB|rp=17.86|mp=83.06}} | ||
<syntaxhighlight lang="java"> | <syntaxhighlight lang="java"> | ||
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int next = i + j; | int next = i + j; | ||
if(next >= dp.length) break; | if(next >= dp.length) break; | ||
if(next == nums.length -1) return true; // reach end already | |||
dp[next] = true; | dp[next] = true; | ||
} | } | ||
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} | } | ||
</syntaxhighlight> | </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; | |||
} | |||
} | |||
</syntaxhighlight> | |||
[[Category:Algorithm]] | |||
[[Category:Dynamic Programming]] | |||
[[Category:Greedy]] | |||
[[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;
}
}