力扣 403 青蛙过河 搜索 || 动态规划

题目链接

代码参考自

    public static boolean flag = false;

    // 用来判断当前走过来距离是否有这样一个数
    public int judge(int []stones, int value) {
        int left = 0, right = stones.length - 1, mid = 0;

        while (left <= right) {
            mid = (left + right) >> 1;

            if (stones[mid] > value) {
                right = mid - 1;
            } else if (stones[mid] < value) {
                left = mid + 1;
            } else {
                return mid;
            }
        }

        return -1;
    }

    // 当前下标，走过来的k值，数组
    public void dfs(int index, int k, HashMap<Integer, HashSet<Integer>> mp, int []stones) {
        if (index >= stones.length - 1 || flag) {
            flag = true;
            return;
        }

        // 减枝操作，如果之前有通过k值走过来，那么没有必要接着往下走了
        // 并且stones[index] + k 在数组有这个值
        int temp = judge(stones, k + stones[index]);
        if (!mp.get(index).contains(k) && temp > -1) {
            mp.get(index).add(k);

            dfs(temp, k - 1, mp, stones);
            dfs(temp, k, mp, stones);
            dfs(temp, k + 1, mp, stones);
        }
    }
    public boolean canCross(int[] stones) {
        if (stones == null || stones.length < 2) {
            return false;
        }

        flag = false;

        HashMap<Integer, HashSet<Integer>> mp = new HashMap<>();
        for (int i = 0; i < stones.length; i++) {
            mp.put(i, new HashSet<>());
        }

        dfs(0, 1, mp, stones);

        return flag;
    }