leetcode 656. Coin Path
Given an array A
(index starts at 1
) consisting of N integers: A1, A2, ..., AN and an integer B
. The integer B
denotes that from any place (suppose the index is i
) in the array A
, you can jump to any one of the place in the array A
indexed i+1
, i+2
, …, i+B
if this place can be jumped to. Also, if you step on the index i
, you have to pay Ai coins. If Ai is -1, it means you can’t jump to the place indexed i
in the array.
Now, you start from the place indexed 1
in the array A
, and your aim is to reach the place indexed N
using the minimum coins. You need to return the path of indexes (starting from 1 to N) in the array you should take to get to the place indexed N
using minimum coins.
If there are multiple paths with the same cost, return the lexicographically smallest such path.
If it's not possible to reach the place indexed N then you need to return an empty array.
Example 1:
Input: [1,2,4,-1,2], 2
Output: [1,3,5]
Example 2:
Input: [1,2,4,-1,2], 1
Output: []
Note:
- Path Pa1, Pa2, ..., Pan is lexicographically smaller than Pb1, Pb2, ..., Pbm, if and only if at the first
i
where Pai and Pbi differ, Pai < Pbi; when no suchi
exists, thenn
<m
. - A1 >= 0. A2, ..., AN (if exist) will in the range of [-1, 100].
- Length of A is in the range of [1, 1000].
- B is in the range of [1, 100].
dp 输出路径
class Solution {
public:
const int inf = 0x3f3f3f3f;
vector<int> cheapestJump(vector<int>& A, int B) {
vector<int> v;
int n = A.size();
if (n == || A[n - ] == -) return v;
vector<int> dp(n, inf);
vector<int> pre(n, -);
dp[n - ] = A[n - ];
for (int i = n - ; i >= ; --i) {
if (A[i] == -) continue;
for (int j = ; j <= B; ++j) {
if (i + j < n && dp[i + j] + A[i] < dp[i]) {
dp[i] = dp[i + j] + A[i];
pre[i] = i + j;
}
}
}
int m = ;
if (dp[] == inf ) return v;
while (m != -) {
v.push_back(m + );
m = pre[m];
}
return v;
}
};
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