数组与矩阵

阳光穿透心脏的1/2处 2021-12-07 13:43 239阅读 0赞

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zODIwMTkzNg_size_16_color_FFFFFF_t_70][]

public void moveZeroes(int[] nums) {
        int idx = 0;
        for (int num : nums) {
            if (num != 0) {
                nums[idx++] = num;
            }
        }
        while (idx < nums.length) {
            nums[idx++] = 0;
        }
    }

# 2. 改变矩阵维度
    
    [566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
    
    ```html
    Input:
    nums =
    [[1,2],
     [3,4]]
    r = 1, c = 4
    
    Output:
    [[1,2,3,4]]
    
    Explanation:
    The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix,
     fill it row by row by using the previous list.
    ```

public int[][] matrixReshape(int[][] nums, int r, int c) {
        int m = nums.length, n = nums[0].length;
        if (m * n != r * c) {
            return nums;
        }
        int[][] reshapedNums = new int[r][c];
        int index = 0;
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                reshapedNums[i][j] = nums[index / n][index % n];
                index++;
            }
        }
        return reshapedNums;
    }

# 3. 找出数组中最长的连续 1
    
    [485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)

public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0, cur = 0;
        for (int x : nums) {
            cur = x == 0 ? 0 : cur + 1;
            max = Math.max(max, cur);
        }
        return max;
    }

# 4. 有序矩阵查找
    
    [240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
    
    ```html
    [
       [ 1,  5,  9],
       [10, 11, 13],
       [12, 13, 15]
    ]

public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        int m = matrix.length, n = matrix[0].length;
        int row = 0, col = n - 1;
        while (row < m && col >= 0) {
            if (target == matrix[row][col]) return true;
            else if (target < matrix[row][col]) col--;
            else row++;
        }
        return false;
    }

# 5. 有序矩阵的 Kth Element
    
    [378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com
    /problems/kth-smallest-element-in-a-sorted-matrix/description/)
    
    ```html
    matrix = [
      [ 1,  5,  9],
      [10, 11, 13],
      [12, 13, 15]
    ],
    k = 8,
    
    return 13.
    ```
    
    解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com
    /problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
    
    二分查找解法：

public int kthSmallest(int[][] matrix, int k) {
        int m = matrix.length, n = matrix[0].length;
        int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            int cnt = 0;
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
                    cnt++;
                }
            }
            if (cnt < k) lo = mid + 1;
            else hi = mid - 1;
        }
        return lo;
    }

堆解法：

public int kthSmallest(int[][] matrix, int k) {
        int m = matrix.length, n = matrix[0].length;
        PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
        for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
        for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
            Tuple t = pq.poll();
            if(t.x == m - 1) continue;
            pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
        }
        return pq.poll().val;
    }
    
    class Tuple implements Comparable<Tuple> {
        int x, y, val;
        public Tuple(int x, int y, int val) {
            this.x = x; this.y = y; this.val = val;
        }
    
        @Override
        public int compareTo(Tuple that) {
            return this.val - that.val;
        }
    }

# 6. 一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数
    
    [645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
    
    ```html
    Input: nums = [1,2,2,4]
    Output: [2,3]
    ```
    
    ```html
    Input: nums = [1,2,2,4]
    Output: [2,3]
    ```
    
    最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、
    O(1) 空间复杂度来求解。
    
    主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。

public int[] findErrorNums(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
                swap(nums, i, nums[i] - 1);
            }
        }
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i + 1) {
                return new int[]{nums[i], i + 1};
            }
        }
        return null;
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }

类似题目：
    
    - [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
    - [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/
    problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
    
    # 7. 找出数组中重复的数，数组值在 [1, n] 之间
    
    [287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/
    find-the-duplicate-number/description/)
    
    要求不能修改数组，也不能使用额外的空间。
    
    二分查找解法：

public int findDuplicate(int[] nums) {
         int l = 1, h = nums.length - 1;
         while (l <= h) {
             int mid = l + (h - l) / 2;
             int cnt = 0;
             for (int i = 0; i < nums.length; i++) {
                 if (nums[i] <= mid) cnt++;
             }
             if (cnt > mid) h = mid - 1;
             else l = mid + 1;
         }
         return l;
    }

双指针解法，类似于有环链表中找出环的入口：

public int findDuplicate(int[] nums) {
        int slow = nums[0], fast = nums[nums[0]];
        while (slow != fast) {
            slow = nums[slow];
            fast = nums[nums[fast]];
        }
        fast = 0;
        while (slow != fast) {
            slow = nums[slow];
            fast = nums[fast];
        }
        return slow;
    }

# 8. 数组相邻差值的个数
    
    [667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems
    /beautiful-arrangement-ii/description/)
    
    ```html
    Input: n = 3, k = 2
    Output: [1, 3, 2]
    Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, 
    and the [2, 1] has exactly 2 distinct integers: 1 and 2.
    ```
    
    题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
    
    让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.

public int[] constructArray(int n, int k) {
        int[] ret = new int[n];
        ret[0] = 1;
        for (int i = 1, interval = k; i <= k; i++, interval--) {
            ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
        }
        for (int i = k + 1; i < n; i++) {
            ret[i] = i + 1;
        }
        return ret;
    }

# 9. 数组的度
    
    [697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
    
    ```html
    Input: [1,2,2,3,1,4,2]
    Output: 6
    ```
    
    题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，
    这个子数组的度和原数组一样。

public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> numsCnt = new HashMap<>();
        Map<Integer, Integer> numsLastIndex = new HashMap<>();
        Map<Integer, Integer> numsFirstIndex = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
            numsLastIndex.put(num, i);
            if (!numsFirstIndex.containsKey(num)) {
                numsFirstIndex.put(num, i);
            }
        }
        int maxCnt = 0;
        for (int num : nums) {
            maxCnt = Math.max(maxCnt, numsCnt.get(num));
        }
        int ret = nums.length;
        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            int cnt = numsCnt.get(num);
            if (cnt != maxCnt) continue;
            ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
        }
        return ret;
    }

# 10. 对角元素相等的矩阵
    
    [766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
    
    ```html
    1234
    5123
    9512
    
    In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", 
    "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements 
    are the same, so the answer is True.

public boolean isToeplitzMatrix(int[][] matrix) {
        for (int i = 0; i < matrix[0].length; i++) {
            if (!check(matrix, matrix[0][i], 0, i)) {
                return false;
            }
        }
        for (int i = 0; i < matrix.length; i++) {
            if (!check(matrix, matrix[i][0], i, 0)) {
                return false;
            }
        }
        return true;
    }
    
    private boolean check(int[][] matrix, int expectValue, int row, int col) {
        if (row >= matrix.length || col >= matrix[0].length) {
            return true;
        }
        if (matrix[row][col] != expectValue) {
            return false;
        }
        return check(matrix, expectValue, row + 1, col + 1);
    }

# 11. 嵌套数组
    
    [565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
    
    ```html
    Input: A = [5,4,0,3,1,6,2]
    Output: 4
    Explanation:
    A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
    
    One of the longest S[K]:
    S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
    ```
    
    题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。

public int arrayNesting(int[] nums) {
        int max = 0;
        for (int i = 0; i < nums.length; i++) {
            int cnt = 0;
            for (int j = i; nums[j] != -1; ) {
                cnt++;
                int t = nums[j];
                nums[j] = -1; // 标记该位置已经被访问
                j = t;
    
            }
            max = Math.max(max, cnt);
        }
        return max;
    }

# 12. 分隔数组
    
    [769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems
    /max-chunks-to-make-sorted/description/)
    
    ```html
    Input: arr = [1,0,2,3,4]
    Output: 4
    Explanation:
    We can split into two chunks, such as [1, 0], [2, 3, 4].
    However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
    ```
    
    题目描述：分隔数组，使得对每部分排序后数组就为有序。

public int maxChunksToSorted(int[] arr) {
        if (arr == null) return 0;
        int ret = 0;
        int right = arr[0];
        for (int i = 0; i < arr.length; i++) {
            right = Math.max(right, arr[i]);
            if (right == i) ret++;
        }
        return ret;
    }

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