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Diffstat (limited to 'other/frog_jump.cpp')
-rw-r--r-- | other/frog_jump.cpp | 47 |
1 files changed, 0 insertions, 47 deletions
diff --git a/other/frog_jump.cpp b/other/frog_jump.cpp deleted file mode 100644 index d9a21fc..0000000 --- a/other/frog_jump.cpp +++ /dev/null @@ -1,47 +0,0 @@ -// 403. Frog Jump -#include "leetcode.h" - -/* - * a frog is crossing a river. the river is divided into some number of units - * and at each unit, there may or may not exist a stone. the frog can jump on a - * stone, but it must not jump into the water. given a list of 'stones' - * positions in (units) in sorted ascending order, determine if the frog can - * cross the river by landing on the last stone. initially, the frog is on the - * first stone and assumes the frist just must be 1 unit. if the frog's last - * jump was k units, its next jujp must be either 'k - 1' or 'k + 1' units. the - * frog can only jump in the forward direction. - */ - -class Solution { - bool process(vector<int> &stones, map<int, int> &mp, int i, int k, - vvd & dp) { - if (i == stones.size() - 1) - return dp[i][k] = 1; - if (dp[i][k] != -1) - return dp[i][k]; - for (int t = k - 1; t <= k + 1; t++) - if (t && mp.find(stones[i] + t) != mp.end()) { - int idx = mp[stones[i] + t]; - if (process(stones, mp, idx, t, dp)) - return dp[i][k] = 1; - } - return dp[i][k] = 0; - } - -public: - bool canCross(vector<int> &stones) { - int n = stones.size(); - map<int, int> mp; - for (int i = 0; i < n; i++) - mp[stones[i]] = i; - vvd dp(n + 1, vector<int>(n + 1, -1)); - return process(stones, mp, 0, 0, dp); - } -}; - -int main() { - Solution obj; - vector<int> s1 = {0, 1, 3, 5, 6, 8, 12, 17}, s2 = {0, 1, 2, 3, 4, 8, 9, 11}; - printf("%d\n", obj.canCross(s1)); // expect: 1 - printf("%d\n", obj.canCross(s2)); // expect: 0 -} |