Personal Blog
  • 💻Notes for Computer Science
  • Leetcode
    • Array
      • Container with most water
      • 3Sum
      • Next Permutation
      • Valid Sudoku
      • Permutation II
      • Combination Sum
      • Triangle
      • Maximal Square
      • Pairs of Songs with Total Duration Divisible by 60
      • Numbers At Most N Given Digit Set
      • Possible Sum
      • Swap Lex Order
      • Partition Equal Subset Sum
      • Domino and Tromino
      • Numbers At Most N Given Digits
      • Car Pooling
      • Surrounding Regions
      • Min Size Subarray Sum
      • Burst Balloons
      • Jump Game I
      • Jump Game II
      • House Robber II
      • Delete and Earn
      • Word Break
      • Decode Ways
      • Longest Increasing Subsequence
      • Cherry Pickup
      • Rotate Image
    • LinkedList
      • IsListPalindrome
      • Linked List Cycle
      • MergeTwoLinkedList
      • ReverseNodeInKGroup
      • RearrangeLastN
      • Remove Duplicates From Sorted List
      • RemoveKFromList
    • String
      • Generate Parentheses
      • Longest Valid Parentheses
      • Longest Common Subsequence
      • Count and Say
      • Decode String
      • Permutation in String
    • Tree
      • House Robber III
      • Convert Sorted Array to Binary Search Tree
      • Restore Binary Tree
      • Populating Next Right Pointers in Each Node II
      • Subtree of Another Tree
    • Graph
      • All Paths from Source to Target
      • Reorder Routes to Make All Paths Lead to the City Zero
      • Max Points on a Line
  • DBMS
    • DBMS Notes
  • Web App
    • Web Design
    • JavaScript
    • React.js
    • ReactNative
    • Mobile Design
    • Dialogue Flow
  • AnaplanIntern
    • Splunk
    • Docker
    • Kubernetes
  • 💰 Notes for Finance Concept
  • Analysis Concept
    • Volume Spread Analysis
    • Smart Money Concepts
Powered by GitBook
On this page
  • Idea
  • Code
  1. Leetcode
  2. Array

Jump Game I

ID: 55

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.

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.

Idea

Greedy + traversal

Start from the beginning, calculate the max index that can be reached from that index. If final index >= nums.length-1, means that can step out, which means can reach end

For calculating, update max to Math.max(i+nums[i], max)

Code

public boolean canJump(int[] nums) {
    int max = nums[0];
    for(int i = 1; i<=max && i<nums.length; i++){
        max = Math.max(max, nums[i]+i);
    }
    return max>=nums.length-1;
}
PreviousBurst BalloonsNextJump Game II

Last updated 3 years ago