You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
- During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that
A[i] <= A[j]andA[j]is the smallest possible value. If there are multiple such indexesj, you can only jump to the smallest such indexj. - During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that
A[i] >= A[j]andA[j]is the largest possible value. If there are multiple such indexesj, you can only jump to the smallest such indexj. - (It may be the case that for some index
i,there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15] Output: 2 Explanation: From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more. From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more. From starting index i = 3, we can jump to i = 4, so we've reached the end. From starting index i = 4, we've reached the end already. In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4] Output: 3 Explanation: From starting index i = 0, we make jumps to i = 1, i = 2, i = 3: During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0]. During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3. During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2]. We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good. In a similar manner, we can deduce that: From starting index i = 1, we jump to i = 4, so we reach the end. From starting index i = 2, we jump to i = 3, and then we can't jump anymore. From starting index i = 3, we jump to i = 4, so we reach the end. From starting index i = 4, we are already at the end. In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2] Output: 3 Explanation: We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 200000 <= A[i] < 100000
A:
思路还是很基本的：就是实现起来，混乱了， 花了2个小时， 哎。
基本思路就是：因为每一步都是确定的。 那么1） 找到每个位置的nextBiggerEq index, and nextSmallerQq index. (借助于 sortedMap )
然后就是递归， 每次走一步。 找到确定的下一个位置。看是否到了end.
class Solution {
public: int oddEvenJumps(vector<int>& A) { int n = A.size(); // find the next smallest value that's bigger than current vector<int> oddBigger(n, INT_MIN); // find the next biggest value that's smaller than current vector<int> evenSmaller(n, INT_MIN); map<int, int> M; // value --> index for (int i = n - 1; i >= 0; i--) { int key = A[i]; if (M.find(key) != M.end()) { oddBigger[i] = M[key]; evenSmaller[i] = M[key]; } else { // lower_bound: (if exist) return an iterator to the first element >= key in a sorted sequence // upper_bound: (if exist) return an iterator to the first element > key in a sorted sequence auto lower = M.lower_bound(key); if (lower != M.end()) { oddBigger[i] = lower->second; } if (lower != M.begin()) { lower--; evenSmaller[i] = lower->second; } } M[key] = i;//assign instead of insert, to overwrite duplicate key } vector<vector<int>> nextStep{ oddBigger, evenSmaller }; vector<vector<STATUS>> status(2, vector<STATUS>(n, UNKNOWN)); status[0][n - 1] = status[1][n - 1] = YES; int res = 0; for (int i = 0; i < n; i++) { if (dfs(A, i, nextStep, status, 0)) res++; } return res; } private: enum STATUS { YES, NO, UNKNOWN }; bool dfs(vector<int>& A, int curIndex, const vector<vector<int>>& nextStep, vector<vector<STATUS>>& status, int oddEvenIndex) { if (curIndex == INT_MIN || status[oddEvenIndex][curIndex] == NO) return false; if (status[oddEvenIndex][curIndex] == YES) return true; int nextIndex = nextStep[oddEvenIndex][curIndex]; bool nextResult = dfs(A, nextIndex, nextStep, status, 1 - oddEvenIndex); status[oddEvenIndex][curIndex] = nextResult ? YES : NO; return nextResult; } };
Learned:
<algorithm> 中， std::lower_bound() 找first element >= key
而upper_bound（) 找 first element > key
**************** 进一步思考 **************
我们其实并不需要后面的dfs步骤。 而在计算之前的可以到达的节点的时候，其实我们也可以用其是否已经arrived end 来表示。
class Solution { public: int oddEvenJumps(vector<int>& A) { int n = A.size(); // next smallest value that's bigger than current vector<bool> oddBigger(n, false); oddBigger[n - 1] = true; // next biggest value that's smaller than current vector<bool> evenSmaller(n, false); evenSmaller[n - 1] = true; map<int, int> M; // value --> index int res = 0; for (int i = n - 1; i >= 0; i--) { int key = A[i]; if (M.find(key) != M.end()) { oddBigger[i] = evenSmaller[M[key]]; evenSmaller[i] = oddBigger[M[key]]; }else { // lower_bound: (if exist) return an iterator to the first element >= key in a sorted sequence auto nextIter = M.lower_bound(key); if (nextIter != M.end()) { oddBigger[i] = evenSmaller[nextIter->second]; } if (nextIter != M.begin()) { nextIter--; evenSmaller[i] = oddBigger[nextIter->second]; } } if (oddBigger[i]) { res++; } M[key] = i;//assign instead of insert, to overwrite duplicate key } return res; } };
Mistakes:
if (oddBigger[i]) {
res++;
}一开始放在else{} 里面了。 比较傻 ~~~~~ 哎
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