LeetCode Weekly Contest 220

1694. Reformat Phone Number

通过对字符串长度对3取模进行分析。

public String reformatNumber(String number) {
        number = number.replaceAll("-", "");
        number = number.replaceAll(" ", "");
        int len = number.length();
        StringBuilder str = new StringBuilder();
        if (len % 3 == 0) {
            int index = 1;
            for (int i = 0; i < len; i++) {
                str.append(number.charAt(i));
                if (index % 3 == 0 && i < len - 1) {
                    str.append("-");
                    index = 0;
                }
                index++;
            }
        }
        if (len % 3 == 1) {
            int n = (len / 3);
            int time = (n - 1) * 3;
            int index = 1;
            for (int i = 0; i < time; i++) {
                str.append(number.charAt(i));
                if (index % 3 == 0) {
                    str.append("-");
                    index = 0;
                }
                index++;
            }
            String substring = number.substring(time);
            str.append(substring.substring(0, 2));
            str.append("-");
            str.append(substring.substring(2));
        }
        if (len % 3 == 2) {
            int time = (len / 3) * 3;
            int index = 1;
            for (int i = 0; i < time; i++) {
                str.append(number.charAt(i));
                if (index % 3 == 0) {
                    str.append("-");
                    index = 0;
                }
                index++;
            }
            str.append(number.substring(time));
        }
        return str.toString();
    }

1695. Maximum Erasure Value

类似最长不重复子序列的，这里是求和最大。

public int maximumUniqueSubarray(int[] nums) {
        int[] q = new int[100005];
        int[] sum = new int[nums.length + 1];
        for (int i = 1; i <= nums.length; i++) {
            sum[i] = sum[i - 1] + nums[i - 1];
        }
        int res = 0;
        int j = 0;
        for (int i = 0; i < nums.length; i++) {
            q[nums[i]]++;
            while (q[nums[i]] > 1) {
                q[nums[j]]--;
                j++;
            }
            res = Math.max(sum[i + 1] - sum[j], res);
        }
        return res;
    }

1696. Jump Game VI

如果在 i 位置，可以跳跃 【i+1, i+k】的位置，求最大值。这题一直想着使用 dfs 来做，没有写出来，看别人的答案，发现使用滑动窗口可以，就是求出【i+1， i+k】中的最大值。

public int maxResult(int[] nums, int k) {
        if (nums.length == 1) {
            return nums[0];
        }
        int ans = Integer.MAX_VALUE;
        int n = nums.length;
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[1] - a[1]);
        pq.offer(new int[]{n - 1, nums[n - 1]});
        for (int i = n - 2; i >= 0; i--) {
            while (pq.peek()[0] > i + k) {
                pq.poll();
            }
            ans = nums[i] + pq.peek()[1];
            pq.offer(new int[]{i, ans});
        }
        return ans;
    }