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Diffstat (limited to 'other/frog_jump.cpp')
-rw-r--r-- | other/frog_jump.cpp | 47 |
1 files changed, 47 insertions, 0 deletions
diff --git a/other/frog_jump.cpp b/other/frog_jump.cpp new file mode 100644 index 0000000..d9a21fc --- /dev/null +++ b/other/frog_jump.cpp @@ -0,0 +1,47 @@ +// 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 +} |