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数组是一种连续存储线性结构，元素类型相同，大小相等，数组是多维的，通过使用整型索引值来访问他们的元素，数组尺寸不能改变。

• 知识点
• 数组与矩阵相关题目

# 知识点

数组的优点:

• 存取速度快

数组的缺点:

• 事先必须知道数组的长度
• 插入删除元素很慢
• 空间通常是有限制的
• 需要大块连续的内存块
• 插入删除元素的效率很低

# 数组与矩阵相关题目

把数组中的 0 移到末尾

283. Move Zeroes (Easy)在新窗口打开

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;}
}

改变矩阵维度

566. Reshape the Matrix (Easy)在新窗口打开

Input:
nums =
[[1,2],[3,4]]
r = 1, c = 4Output:
[[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;
}

找出数组中最长的连续 1

485. Max Consecutive Ones (Easy)在新窗口打开

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;
}

有序矩阵查找

240. Search a 2D Matrix II (Medium)在新窗口打开

[[ 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;
}

有序矩阵的 Kth Element

378. Kth Smallest Element in a Sorted Matrix ((Medium))在新窗口打开

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在新窗口打开

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;}@Overridepublic int compareTo(Tuple that) {return this.val - that.val;}
}

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)在新窗口打开

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;
}

类似题目:

• 448. Find All Numbers Disappeared in an Array (Easy)在新窗口打开，寻找所有丢失的元素
• 442. Find All Duplicates in an Array (Medium)在新窗口打开，寻找所有重复的元素。

找出数组中重复的数，数组值在 [1, n] 之间

287. Find the Duplicate Number (Medium)在新窗口打开

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;
}

数组相邻差值的个数

667. Beautiful Arrangement II (Medium)在新窗口打开

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;
}

数组的度

697. Degree of an Array (Easy)在新窗口打开

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;
}

对角元素相等的矩阵

766. Toeplitz Matrix (Easy)在新窗口打开

1234
5123
9512In 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);
}

嵌套数组

565. Array Nesting (Medium)在新窗口打开

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;
}

分隔数组

769. Max Chunks To Make Sorted (Medium)在新窗口打开

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;
}