zoukankan      html  css  js  c++  java
  • LeetCode

    题目:

    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.)

    Subscribe to see which companies asked this question

    思路:

    1)用一个dp数组来记录到达当前位置的最小步数

    package dp;
    
    public class JumpGameII {
    
        public int jump(int[] nums) {        
            int len = nums.length;
            int[] dp = new int[len];
            for (int i = 0; i < len; ++i)
                dp[i] = Integer.MAX_VALUE;
            dp[0] = 0;
            // In order to save time, we don't visit the elements which already been visited.
            int start = 1;
            for (int i = 0; i < len - 1; ++i) {
                start = go(i, nums[i], start, dp);
            }
            
            return dp[len-1];
        }
        
        // Get next start position.
        private int go(int current, int step, int start, int[] dp) {
            int i = start;
            for (; i <= current + step && i < dp.length; ++i) {
                if (dp[i] > dp[current] + 1)
                    dp[i] = dp[current] + 1;
            }
            
            return i;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            int[] nums = { 2,3,1,1,4 };
            JumpGameII j = new JumpGameII();
            System.out.println(j.jump(nums));
        }
    
    }

    2)不用数组,就直接往前移动,第一次扫走一步能到的地方,第二次扫走二步能到的地方,在这个过程中,只要碰到了最后一个位置,就终止,证明是最小步数。

    package dp;
    
    public class JumpGameII {
    
        public int jump(int[] nums) {
            int len; 
            if ((len = nums.length) == 1) return 0;
            int count = 0;
            int start = 0;
            int end = 0;
            while (end < len) {
                int mostRight = 0;
                ++count;
                for (int j = start; j <= end; ++j) {
                    if (j + nums[j] >= len - 1)
                        return count;
                    if (j + nums[j] > mostRight)
                        mostRight = j + nums[j];
                }
                
                start  = end + 1;
                end = mostRight;
            }
            
            return count;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            int[] nums = { 2,3,1,1,4 };
            JumpGameII j = new JumpGameII();
            System.out.println(j.jump(nums));
        }
    
    }
  • 相关阅读:
    SAP系统报错
    基金投资
    Reading: 重构相关
    C/C++: static variables
    C/C++: 如何删除本地文件/读取某个目录下符合某种pattern的所有文件路径
    如何在Linux下建立包含lua vm的unit test framwork
    C++实现字符串分割(类似于Python的split方法)
    postMan 汉化
    搞懂MySQL InnoDB事务ACID实现原理
    mysql 索引优化
  • 原文地址:https://www.cnblogs.com/null00/p/5022937.html
Copyright © 2011-2022 走看看