[leetcode]45. Jump Game II

链接：https://leetcode.com/problems/jump-game-ii/description/

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.

方法一：动态规划（超时）

class Solution {
public:
    int jump(vector<int>& nums) {
        vector<int> dp(nums.size(),nums.size());
        dp[0]=0;
        for(int i=0;i<nums.size();i++)
        {
            for(int step=nums[i];step>0;step--)
            {
                if(step+i<nums.size())
                {
                    dp[step+i]=min(dp[i]+1,dp[step+i]);
                }
            }
        }
        return dp[nums.size()-1];
    }
};

方法二：（贪心）

这道题，需要求的是最少的步数。因此需要添加step变量记录最少步数。至于什么时候step需要加1？答案是当前的i超过了前一步的最远位置。所以引入last变量记录上一步能到达的最远位置。reach、step、last的初始值均为0。当遍历到i的时候，如果i超过了last（即上一步能到达的最远位置），说明步数需要加1（即step++），此时仍需要更新last为当前最远位置reach。全程只需遍历1次数组，而且空间复杂度为常量。

class Solution {
public:
    int jump(vector<int>& nums) {
        int reach = 0, last = 0, step = 0,n=nums.size();
        for(int i = 0; i <= reach && i < n; ++ i)
        {
            if(i > last)
            {
                ++ step;
                last = reach;
            }
            reach = max(reach, nums[i] + i);
        }
        return reach >= n - 1 ? step : 0;
    }
};