Leetcode 题解 - 数组与矩阵

1. 把数组中的 0 移到末尾

1. Move Zeroes (Easy)

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For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

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

1. Reshape the Matrix (Easy)

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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

1. Max Consecutive Ones (Easy)

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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

1. Search a 2D Matrix II (Medium)

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[
   [ 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

1. Kth Smallest Element in a Sorted Matrix ((Medium))

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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

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

1. Set Mismatch (Easy)

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Input: nums = [1,2,2,4]
Output: [2,3]

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;
}

7. 找出数组中重复的数，数组值在 [1, n] 之间

1. Find the Duplicate Number (Medium)

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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

1. Beautiful Arrangement II (Medium)

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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

1. Degree of an Array (Easy)

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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

1. Toeplitz Matrix (Easy)

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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

1. Array Nesting (Medium)

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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

1. Max Chunks To Make Sorted (Medium)

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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;
}