1. 移动零
Leetcode (opens new window)/ 力扣
class Solution {
// 双指针
// j 从左至右找不为 0 的数
// i 负责维护数组的边界
public void moveZeroes(int[] nums) {
int i = 0;
int j = 0;
int n = nums.length;
while (j < n) {
if (nums[j] != 0) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
i++;
}
j++;
}
}
}
class Solution {
public void moveZeroes(int[] nums) {
int j = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0) {
nums[j++] = nums[i];
}
}
for (int i = j; i < nums.length; i++) {
nums[i] = 0;
}
}
}
2. 重塑矩阵
Leetcode (opens new window)/ 力扣
class Solution {
public int[][] matrixReshape(int[][] mat, int r, int c) {
int matR = mat.length;
int matC = mat[0].length;
if (matC * matR != r * c) {
return mat;
}
int[][] result = new int[r][c];
int index = 0;
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
result[i][j] = mat[index / matC][index % matC];
index++;
}
}
return result;
}
}
3. 最大连续 1 的个数
Leetcode (opens new window)/ 力扣
class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int max = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 1) {
int value = map.getOrDefault(i - 1, 0) + 1;
if (value > max) {
max = value;
}
map.put(i, value);
}
}
return max;
}
}
进行优化：
class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int last = 0;
int max = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 1) {
last = last + 1;
if (last > max) {
max = last;
}
} else {
last = 0;
}
}
return max;
}
}
4. 搜索二维矩阵 II
Leetcode (opens new window)/ 力扣
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int i = matrix.length - 1;
int j = 0;
// 从左下角开始找
while (i >= 0 && j < matrix[i].length) {
if (matrix[i][j] == target) {
return true;
}
// 如果大于就向上一行去找
if (matrix[i][j] > target) {
i--;
// 如果小于就向左一列去找
} else if (matrix[i][j] < target) {
j++;
}
}
return false;
}
}
5. 有序矩阵的 Kth Element *
Leetcode (opens new window)/ 力扣
6. 错误的集合
Leetcode (opens new window)/ 力扣
class Solution {
public int[] findErrorNums(int[] nums) {
Map<Integer, Integer> map = new HashMap<>();
int n = nums.length;
for (int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
}
int[] result = new int[2];
for (int i = 0; i <= n; i++) {
int count = map.getOrDefault(i, 0);
if (count == 2) {
result[0] = i;
}
if (count == 0) {
result[1] = i;
}
}
return result;
}
}
7. 寻找重复数
Leetcode (opens new window)/ 力扣(opens new window)
要求不能修改数组，也不能使用额外的空间。
双指针法：
class Solution {
public int findDuplicate(int[] nums) {
int slow = 0;
int fast = 0;
slow = nums[slow];
fast = nums[nums[fast]];
// 进入环
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
int pre1 = 0;
int pre2 = slow;
// 找环的入口
while (pre1 != pre2) {
pre1 = nums[pre1];
pre2 = nums[pre2];
}
return pre1;
}
}
8. 优美的排列 II
Leetcode (opens new window)/ 力扣
class Solution {
public int[] constructArray(int n, int k) {
int[] result = new int[n];
// 构造一部分等差数列
for (int i = 0; i < n - k - 1; i++) {
result[i] = i + 1;
}
int left = n - k;
int right = n;
int j = 0;
// 构造一部分交错数列
for (int i = n - k - 1; i < n; i++) {
if (j % 2 == 0) {
result[i] = left;
left++;
} else {
result[i] = right;
right--;
}
j++;
}
return result;
}
}
9. 数组的度
Leetcode (opens new window)/ 力扣
class Solution {
public int findShortestSubArray(int[] nums) {
// 元素为 key，值为一个 int 类型的数组
// 数组：[0] 为 出现的次数，[1] 为第一次出现的位置
// [2] 为最后一次出现的位置
Map<Integer, int[]> map = new HashMap<>();
// 存入哈希表
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i])) {
map.get(nums[i])[0]++;
map.get(nums[i])[2] = i;
} else {
int[] value = new int[3];
value[0] = 1;
value[1] = i;
value[2] = i;
map.put(nums[i], value);
}
}
int maxNum = 0;
int minSize = 0;
for (Map.Entry<Integer, int[]> entry : map.entrySet()) {
int[] value = entry.getValue();
// 如果小于说明这才是数组的度
if (maxNum < value[0]) {
maxNum = value[0];
minSize = value[2] - value[1] + 1;
// 当度相同时，判断谁的连续子数组的的最短，将其返回
} else if (maxNum == value[0]) {
minSize = Math.min(minSize, value[2] - value[1] + 1);
}
}
return minSize;
}
}
10. 对角元素相等的矩阵
Leetcode (opens new window)/ 力扣
class Solution {
public boolean isToeplitzMatrix(int[][] matrix) {
if (matrix == null) {
return false;
}
for (int i = 0; i < matrix.length - 1; i++) {
for (int j = 0; j < matrix[i].length -1; j++) {
if (matrix[i][j] != matrix[i + 1][j + 1]) {
return false;
}
}
}
return true;
}
}
11. 数组嵌套 *
Leetcode (opens new window)/ 力扣
class Solution {
public int arrayNesting(int[] nums) {
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != Integer.MAX_VALUE) {
int start = nums[i];
int count = 0;
while (nums[start] != Integer.MAX_VALUE) {
int temp = start;
start = nums[start];
count++;
nums[temp] = Integer.MAX_VALUE;
}
result = Math.max(result, count);
}
}
return result;
}
}
12. 最多能完成排序的块
Leetcode (opens new window)/ 力扣
class Solution {
public int maxChunksToSorted(int[] arr) {
int max = 0;
int result = 0;
for (int i = 0; i < arr.length; i++) {
max = Math.max(max, arr[i]);
// 如果某一段的最大值等于当前下标
// 说明这一段k
if (max == i) {
result++;
}
}
return result;
}
}
