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-rw-r--r--other/frog_jump.cpp47
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diff --git a/other/frog_jump.cpp b/other/frog_jump.cpp
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-// 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
-}