• Jump Game


    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.

    Example:

    Input: [2,3,1,1,4]
    Output: 2
    Explanation: 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.

    class Solution {
    public:
        int jump(vector<int>& nums) {
            int len = nums.size(),start = 0,end = 0, step = 0;
            while (end < len - 1)
            {
                ++step;
                int maxEnd = end + 1;
                for (int i = start; i <= end; ++i)
                {
                    if (nums[i] + i > len - 1) return step;
                    maxEnd = max(nums[i] + i,maxEnd);
                }
                start = end + 1;
                end = maxEnd;
            }
            return step;
        }
    };

    2.

    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.

    Example 1:

    Input: [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: [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.
    // trace back
    class Solution {
    public:
        bool canJump(vector<int>& nums) {
            int len = nums.size();
            if (len == 1)return true;
            int max_index = 0;
            for (int i = 0; i <= max_index;++i)
            {
                if (nums[i] + i >= len - 1)return true;
                if (max_index < i)return false;
                max_index = max(max_index,nums[i] + i);
            }
            return false;
        }
    };
  • 相关阅读:
    Linux 环境下umount, 报 device is busy 的问题分析与解决方法
    WScript与CScript的区别
    20170803上课笔记
    20170802上课随笔
    20170801上课笔记
    20170731上课笔记
    20170729上课笔记
    20170727上课笔记
    20170726上课笔记
    20170725上课笔记
  • 原文地址:https://www.cnblogs.com/qiang-wei/p/11908910.html
Copyright © 2020-2023  润新知