[Leetcode]Jump Game 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.

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.

class Solution {
public:
    /*algorithm  DP
        dp(i): minimum steps to end from ith position
        dp(n-1) = 0
        dp(i) = min{distance(i,j) + dp(j)} i+1<=j<n , distance(i,j) <= A[i]
        TME error
  */
    int jump(vector<int>& nums) {
        int n = nums.size();
        vector<int>dp(n,INT_MAX);
        dp[n-1]=0;
        for(int i = n-2;i>=0;i--){
            for(int j = 1;j < n-i;j++){
                if(nums[i] >= j){
                    dp[i] = min(dp[i],1+dp[i+j]);
                }
            }
        }
        return dp[0]==INT_MAX?-1:dp[0];
    }
};

class Solution {
public:
    /*algorithm : greedy
        prev: last max reachable position
        cur: current max reachable position
        cur: max(cur,i+nums[i)
        [3,1,1]
    */
    int jump(vector<int>& nums) {
        int n = nums.size();
        int prev = 0,cur = 0,jumps=0;
        for(int i = 0;i < n;i++){
            if(i > prev){
                if(cur==prev)return -1;
                prev = cur;
                jumps++;
            }
            cur = max(cur,i+nums[i]);
        }
        return jumps;
    }
};