zoukankan      html  css  js  c++  java
  • 55. Jump Game I && II

    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.

    For example:
    A = [2,3,1,1,4], return true.

    A = [3,2,1,0,4], return false.

    设定一个上线reach 在每个值上;

    public class Solution {
        public boolean canJump(int[] nums) {
            int index = 0;
            int reach = 0;
            while(index < nums.length && index <= reach){
                reach = Math.max(reach, nums[index] + index);
                index ++;
            }
            return reach >= nums.length -1;
        }
    }

    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.

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

    For example:
    Given array A = [2,3,1,1,4]

    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.)

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

    public class Solution {
        public int jump(int[] nums) {
            int reach = 0;
            int index = 0;
            int max = nums[0];
            for(int i = 0; i < nums.length ; i++){
                if(reach < i){
                    index++;
                    reach = max;
                }
                max = Math.max(max, nums[i] + i);
            }
            return index;
        }
    }
  • 相关阅读:
    平衡二叉树之RB树
    平衡二叉树之AVL树
    实现哈希表
    LeetCode Median of Two Sorted Arrays
    LeetCode Minimum Window Substring
    LeetCode Interleaving String
    LeetCode Regular Expression Matching
    PAT 1087 All Roads Lead to Rome
    PAT 1086 Tree Traversals Again
    LeetCode Longest Palindromic Substring
  • 原文地址:https://www.cnblogs.com/joannacode/p/5958894.html
Copyright © 2011-2022 走看看