Q****540. Single Element in a Sorted Array
You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once. Find this single element that appears only once.
Follow up: Your solution should run in O(log n) time and O(1) space.
Example 1:
Input: nums = [1,1,2,3,3,4,4,8,8]
Output: 2
Example 2:
Input: nums = [3,3,7,7,10,11,11]
Output: 10
Constraints:
-
1 <= nums.length <= 10^5 -
0 <= nums[i] <= 10^5
解法及注释
class Solution {
public int singleNonDuplicate(int[] nums) {
if(nums.length == 1)
return nums[0];
int left = 0, right = nums.length - 1;
if(nums[left] != nums[left + 1])
return nums[left];
else if(nums[right] != nums[right - 1])
return nums[right];
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] != nums[mid - 1] && nums[mid] != nums[mid + 1])
return nums[mid];
else if(mid % 2 == 0 && nums[mid] == nums[mid + 1] ||
mid % 2 == 1 && nums[mid] == nums[mid - 1])
left = mid + 1;
else
right = mid - 1;
}
return -1;
}
}
Q****153. Find Minimum in Rotated Sorted Array
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
[4,5,6,7,0,1,2]if it was rotated4times.[0,1,2,4,5,6,7]if it was rotated7times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums, return the minimum element of this array.
Example 1:
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000- All the integers of
numsare unique. numsis sorted and rotated between1andntimes.
解法及注释
class Solution {
public int findMin(int[] nums) {
if(nums.length == 1)
return nums[0];
int left = 0, right = nums.length - 1;
//case 1: if there is no rotation
if(nums[right] > nums[0])
return nums[0];
//case 2: rotation exists in the array
while(left <= right) {
int mid = left + (right - left)/2;
if(nums[mid] > nums[mid + 1])
return nums[mid + 1];
if(nums[mid - 1] > nums[mid]) {
return nums[mid];
}
if(nums[mid] > nums[0]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
}
Q****154. Find Minimum in Rotated Sorted Array II
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
Find the minimum element.
The array may contain duplicates.
Example 1:
Input: [1,3,5]
Output: 1
Example 2:
Input: [2,2,2,0,1]
Output: 0
Note:
- This is a follow up problem to Find Minimum in Rotated Sorted Array.
- Would allow duplicates affect the run-time complexity? How and why?
解法及注释
class Solution {
public int findMin(int[] nums) {
if(nums.length == 1)
return nums[0];
int left = 0, right = nums.length - 1;
while(left < right) {
int mid = left + (right - left) / 2;
if(nums[mid] < nums[right])
right = mid;
else if(nums[mid] > nums[right])
left = mid + 1;
else if(nums[left] < nums[right])
return nums[left];
else {
left++;
}
}
return nums[left];
}
}
Q****74. Search a 2D Matrix
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
- Integers in each row are sorted from left to right.
- The first integer of each row is greater than the last integer of the previous row.
Example 1:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true
Example 2:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false
Constraints:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 100-104 <= matrix[i][j], target <= 104
解法及注释
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int i = 0, j = matrix[0].length - 1;
while(i >=0 && i < matrix.length && j >=0 && j < matrix[0].length) {
if(matrix[i][j] == target)
return true;
else if(matrix[i][j] > target) {
j--;
} else if(matrix[i][j] < target) {
i++;
}
}
return false;
}
}
Q****240. Search a 2D Matrix II
Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:
- Integers in each row are sorted in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
Example 1:
Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true
Example 2:
Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false
Constraints:
m == matrix.lengthn == matrix[i].length1 <= n, m <= 300-109 <= matix[i][j] <= 109- All the integers in each row are sorted in ascending order.
- All the integers in each column are sorted in ascending order.
-109 <= target <= 109
解法及注释
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix.length == 1 && matrix[0].length == 1)
return matrix[0][0] == target;
int row = matrix[0].length, col = matrix.length;
for(int i = 0; i < col; i++) {
if(matrix[i][0] <= target && matrix[i][row - 1] >= target) {
for(int j = 0; j < row; j++) {
if(matrix[i][j] == target)
return true;
}
} else
continue;
}
return false;
}
}
Q****378. Kth Smallest Element in a Sorted Matrix
Given an n x n matrix where each of the rows and columns are sorted in ascending order, return the kth smallest element in the matrix.
Note that it is the kth smallest element in the sorted order, not the kth distinct element.
Example 1:
Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15],
and the 8th largest number is 13
Example 2:
Input: matrix = [[-5]], k = 1
Output: -5
Constraints:
n == matrix.lengthn == matrix[i].length1 <= n <= 300-109 <= matrix[i][j] <= -109- All the rows and columns of
matrixare guaranteed to be sorted in non-degreasing order. 1 <= k <= n2
解法及注释
class Solution {
public int kthSmallest(int[][] matrix, int k) {
if(matrix == null || matrix.length == 0)
return -1;
if(k > matrix.length * matrix[0].length)
return -1;
PriorityQueue<Element> queue = new PriorityQueue<Element>((Element a, Element b) -> (a.val - b.val));
for(int i = 0; i < matrix.length; i++) {
Element element = new Element(matrix[i][0], 0, i);
queue.add(element);
}
while(!queue.isEmpty() && k > 0) {
Element e = queue.peek();
queue.remove(e);
k--;
if(k == 0)
return e.val;
if(e.currIdx < matrix[e.rowIdx].length - 1) {
e.currIdx += 1;
e.val = matrix[e.rowIdx][e.currIdx];
queue.add(e);
}
}
return -1;
}
}
class Element {
int val;
int currIdx;
int rowIdx;
public Element(int val, int currIdx, int rowIdx) {
this.val = val;
this.currIdx = currIdx;
this.rowIdx = rowIdx;
}
}
