每周LeetCode算法题（十五）403. Frog Jump

每周LeetCode算法题（十五）

题目: 403. Frog Jump

A frog is crossing a river. The river is divided into x 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 is able to cross the river by landing on the last stone. Initially, the frog is on the first stone and assume the first jump must be 1 unit.

If the frog’s last jump was k units, then its next jump must be either k - 1, k, or k + 1 units. Note that the frog can only jump in the forward direction.

解法分析

本题需要用递归的方法求解。由于最好能尽快得出“有解”的解，需要深度优先的解法，否则经过我的多次尝试，都是超时，故用递归。青蛙能否从一个位置跳到下一个位置，取决于下一个位置上是否有石头，并且下一个位置还能否继续往前跳。具体代码如下。

C++代码

class Solution {
public:
    unordered_set<int> st;
    bool canCross(vector<int>& stones) {
        if (stones[0] != 0 || stones[1] != 1) {
            return false;
        }
        st.insert(stones[0]);
        for (int i = 1; i < stones.size(); i++) {
            st.insert(stones[i]);
            if (stones[i] - stones[i - 1] > i) {
                return false;
            }
        }
        return can(1, 1, stones[stones.size() - 1]);
    }

    bool can(int pos, int k, int end) {
        if (pos + k <= end + 1 && pos + k >= end - 1) {
            return true;
        }
        if (st.find(pos + k + 1) != st.end() && can(pos + k + 1, k + 1, end)) {
            return true;
        }
        if (st.find(pos + k) != st.end() && can(pos + k, k, end)) {
            return true;
        }
        if (k > 1 && st.find(pos + k - 1) != st.end() && can(pos + k - 1, k - 1, end)) {
            return true;
        }

        return false;
    }
};