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:
    Python笔记 #17# Pandas: Merge
    MVC相关资料收集
    Python笔记 #16# Pandas: Operations
    Least slack time scheduling
    Python笔记 #15# Pandas: Missing Data
    Python笔记 #14# Pandas: Selection
    C++中const引用的是对象的时候只能调用该对象的f()const方法
    模板与泛型编程
    c++中的单例模式
    C/C++异常处理机制
  • 原文地址:https://www.cnblogs.com/null00/p/5022937.html
Copyright © 2011-2022 走看看