Leetcode 第34场双周赛题解

前60名有奖，搞了个62名，还是菜。

5491. 矩阵对角线元素的和

思路：直接求和就行，只是n为奇数的时候中间的数会计算两次，所以要减去一次。

class Solution {
    public int diagonalSum(int[][] mat) {

        int sum = 0;
        int n = mat.length;

        for (int i = 0; i < n; i++)
            sum += mat[i][i];

        for (int i = 0; i < n; i++) {
            if (i == n - 1 - i)
                continue;
            sum += mat[i][n - 1 - i];
        }

        return sum;

    }
}

5492. 分割字符串的方案数

思路：本质上就是找到三段中两个间隔内零的个数，直接相乘即可。当然要特判掉一些特殊情况（比如：全是0或者无法分成三段时）

class Solution {

    static int mod = 1000000007;

    public int numWays(String s) {

        int sum = 0;
        long ans = 0;
        int n = s.length();

        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1')
                sum++;
        }

        if (sum % 3 != 0)
            return 0;

        if (sum == 0) {
            ans = n - 1;
            ans = ans * (ans - 1) / 2 % mod;
            return (int) ans;
        }

        sum /= 3;
        int l = 0, r = n - 1, now = 0;

        for (; l < n && now < sum; l++) {
            if (s.charAt(l) == '1')
                now++;
        }

        now = 0;
        for (; r >= 0 && now < sum; r--) {
            if (s.charAt(r) == '1')
                now++;
        }

        long left = 0, right = 0;
        while (s.charAt(l) == '0') {
            left++;
            l++;
        }
        while (s.charAt(r) == '0') {
            right++;
            r--;
        }

        left++;
        right++;

        //System.out.println(l + " " + r);

        ans = left;
        ans = ans * right % mod;

        return (int) ans;

    }
}

5493. 删除最短的子数组使剩余数组有序

思路：这种类型的题基本就是换换包装，本质考察的知识一点没变，我们找到头和尾分别能组成的最长连续非递减子串，之后暴力左边的长度，然后二分右边的长度，中间的删掉，标记下最优答案即可。

class Solution {
    public int findLengthOfShortestSubarray(int[] arr) {

        int n = arr.length;

        if (n <= 1) return 0;

        int l = 1, r = n - 2;
        while (l < n && arr[l] >= arr[l - 1]) l++;
        while (r >= 0 && arr[r] <= arr[r + 1]) r--;

        if (l == n || r < 0) return 0;

        //System.out.println(l + " " + r);

        int ans = Math.max(l, n - r - 1);

        for (int i = 0; i < l; i++) {
            int ls = r + 1, rs = n - 1, p = n;
            while (ls <= rs) {
                int mid = (ls + rs) / 2;
                if (arr[mid] >= arr[i]) {
                    p = mid;
                    rs = mid - 1;
                } else
                    ls = mid + 1;
            }
            if (p != -1)
                ans = Math.max(ans, i + 1 + n - p);
        }

        return n - ans;

    }
}

5494. 统计所有可行路径

思路：可以考虑采用记忆化搜索，直接统计方案数。【时间复杂度较高】

class Solution {

    private int n;
    private int[] locations;
    private int mod = 1000000007;
    private Map map;

    public int countRoutes(int[] locations, int start, int finish, int fuel) {

        n = locations.length;
        map = new HashMap<>();
        this.locations = locations;

        return dfs(start, finish, fuel);

    }

    private int dfs(int u, int end, int fuel) {

        if (fuel < 0) return 0;

        if (fuel == 0)
            return u == end ? 1 : 0;

        String s = String.valueOf(u) + "#" + String.valueOf(fuel);

        if (map.containsKey(s))
            return map.get(s);

        long res = (u == end ? 1 : 0);

        for (int i = 0; i < n; i++) {
            if (i == u)
                continue;
            res = (res + dfs(i, end, fuel - Math.abs(locations[u] - locations[i]))) % mod;
        }

        map.put(s, (int) res);

        return (int) res;

    }
}