[LeetCode] 656. Coin Path 硬币路径
Given an array A (index starts at 1) consisting of N integers: A1, A2, ..., AN and an integer B. The integer Bdenotes 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
iwhere Pai and Pbi differ, Pai < Pbi; when no suchiexists, 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].
给一个数组A，数组元素的值代表cost，一个整数B表示能走的最大步数。从第1个位置开始走，每次能走的步数是B步以内，走到某个位置就要付出该位置的cost，目标是到达最末尾位置，使得付出总cost值最小，输出所有路径。如果某个位置是-1，不可以走这个位置。如果有多个路径，输出按字母顺序排列。
解法：DP, 从后往前跳。首先判断最后一个位置是否为-1，如果是说明无法到达最后位置，返回空。用一个一维数组dp记录跳到i位置所用的最小cost， 从i位置跳时有B种跳法，再去判断每一个跳法的dp值。
dp[i] = A[i] + dp[j] (dp[j]为从B种跳法中的一种的dp值)
Python:
# Time: O(n * B)
# Space: O(n)
class Solution(object):
def cheapestJump(self, A, B):
"""
:type A: List[int]
:type B: int
:rtype: List[int]
"""
result = []
if not A or A[-1] == -1:
return result
n = len(A)
dp, next_pos = [float("inf")] * n, [-1] * n
dp[n-1] = A[n-1]
for i in reversed(xrange(n-1)):
if A[i] == -1:
continue
for j in xrange(i+1, min(i+B+1,n)):
if A[i] + dp[j] < dp[i]:
dp[i] = A[i] + dp[j]
next_pos[i] = j
if dp[0] == float("inf"):
return result
k = 0
while k != -1:
result.append(k+1)
k = next_pos[k]
return result
C++:
class Solution {
public:
vector<int> cheapestJump(vector<int>& A, int B) {
if (A.back() == -1) return {};
int n = A.size();
vector<int> res, dp(n, INT_MAX), pos(n, -1);
dp[n - 1] = A[n - 1];
for (int i = n - 2; i >= 0; --i) {
if (A[i] == -1) continue;
for (int j = i + 1; j <= min(i + B, n - 1); ++j) {
if (dp[j] == INT_MAX) continue;
if (A[i] + dp[j] < dp[i]) {
dp[i] = A[i] + dp[j];
pos[i] = j;
}
}
}
if (dp[0] == INT_MAX) return res;
for (int cur = 0; cur != -1; cur = pos[cur]) {
res.push_back(cur + 1);
}
return res;
}
};
C++:
class Solution {
public:
vector<int> cheapestJump(vector<int>& A, int B) {
if (A.back() == -1) return {};
int n = A.size();
vector<int> res, dp(n, INT_MAX), pos(n, -1), len(n, 0);
dp[0] = 0;
for (int i = 0; i < n; ++i) {
if (A[i] == -1) continue;
for (int j = max(0, i - B); j < i; ++j) {
if (dp[j] == INT_MAX) continue;
int t = A[i] + dp[j];
if (t < dp[i] || (t == dp[i] && len[i] < len[j] + 1)) {
dp[i] = t;
pos[i] = j;
len[i] = len[j] + 1;
}
}
}
if (dp[n - 1] == INT_MAX) return res;
for (int cur = n - 1; cur != -1; cur = pos[cur]) {
res.insert(res.begin(), cur + 1);
}
return res;
}
};
C++:
class Solution {
public:
vector<int> cheapestJump(vector<int>& A, int B) {
vector<int> result;
if (A.empty() || A.back() == -1) {
return result;
}
const int n = A.size();
vector<int> dp(n, numeric_limits<int>::max()), next(n, -1);
dp[n - 1] = A[n - 1];
for (int i = n - 2; i >= 0; --i) {
if (A[i] == -1) {
continue;
}
for (int j = i + 1; j <= min(i + B, n - 1); ++j) {
if (dp[j] == numeric_limits<int>::max()) {
continue;
}
if (A[i] + dp[j] < dp[i]) {
dp[i] = A[i] + dp[j];
next[i] = j;
}
}
}
if (dp[0] == numeric_limits<int>::max()) {
return result;
}
int k = 0;
while (k != -1) {
result.emplace_back(k + 1);
k = next[k];
}
return result;
}
};
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