wordpress二次开发手册,慈溪企业网站seo,焕识品牌设计,优化大师的功能有哪些双指针
在处理数组和链表相关问题时#xff0c;双指针技巧是经常用到的#xff0c;双指针技巧主要分为两类#xff1a;左右指针和快慢指针。所谓左右指针#xff0c;就是两个指针相向而行或者相背而行#xff1b;而所谓快慢指针#xff0c;就是两个指针同向而行#xf…双指针
在处理数组和链表相关问题时双指针技巧是经常用到的双指针技巧主要分为两类左右指针和快慢指针。所谓左右指针就是两个指针相向而行或者相背而行而所谓快慢指针就是两个指针同向而行一快一慢。
15. 三数之和⭐️⭐️
思路
两数之和 - 三数之和 - N 数之和
代码
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;public class ThreeSumTarget {public ListListInteger threeSum(int[] nums) {ListListInteger res new ArrayList();Arrays.sort(nums);for (int i 0; i nums.length; i) {if (nums[i] 0) {return res;}if (i 0 nums[i] nums[i - 1]) {continue;}int left i 1;int right nums.length - 1;while (left right) {int sum nums[i] nums[left] nums[right];if (sum 0) {left;} else if (sum 0) {right--;} else {res.add(Arrays.asList(nums[i], nums[left], nums[right]));while ((left right) (nums[right] nums[right - 1])) {right--;}while ((left right) nums[left] nums[left 1]) {left;}right--;left}}}return res;}}import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;public class NSumTarget {public ListListInteger threeSum(int[] nums) {Arrays.sort(nums);return nSumTarget(nums, 3, 0, 0);}private ListListInteger nSumTarget(int[] nums, int n, int start, int target) {int size nums.length;ListListInteger res new ArrayList();if (n 2 || size n) {return res;}if (n 2) {int low start;int high size - 1;while (low high) {int sum nums[low] nums[high];int left nums[low];int right nums[high];if (sum target) {while (low high nums[low] left) {low;}} else if (sum target) {while (low high nums[high] right) {high--;}} else {res.add(new ArrayList(Arrays.asList(left, right)));while (low high nums[low] left) {low;}while (low high nums[high] right) {high--;}}}} else {for (int i start; i size; i) {if (i start nums[i] nums[i - 1]) {continue;}ListListInteger sub nSumTarget(nums, n - 1, i 1, target - nums[i]);for (ListInteger arr : sub) {arr.add(nums[i]);res.add(arr);}while (i size - 1 nums[i] nums[i 1]) {i;}}}return res;}}
复杂度
时间复杂度O(N^2)空间复杂度O(logN)
5. 最长回文子串⭐️⭐
思路
寻找回文串的问题核心思想是从中间开始向两边扩散来判断回文串对于最长回文子串就是这个意思
for 0 i len(s):找到以 s[i] 为中心的回文串更新答案
找回文串的关键技巧是传入两个指针 left 和 right 向两边扩散因为这样实现可以同时处理回文串长度为奇数和偶数的情况。
for 0 i len(s):# 找到以 s[i] 为中心的回文串palindrome(s, i, i)# 找到以 s[i] 和 s[i1] 为中心的回文串palindrome(s, i, i 1)更新答案
代码
public class LongestPalindromicSubstring {public String longestPalindrome(String s) {String res ;for (int i 0; i s.length(); i) {String s1 palindrome(s, i, i); // 奇数情况String s2 palindrome(s, i, i 1); // 偶数情况res res.length() s1.length() ? res : s1;res res.length() s2.length() ? res : s2;}return res;}private String palindrome(String s, int left, int right) {while (left 0 right s.length() s.charAt(left) s.charAt(right)) {left--;right;}return s.substring(left 1, right);}}
复杂度
时间复杂度O(N^2)空间复杂度O(N)
88. 合并两个有序数组⭐️⭐️
思路
代码
public class MergeTwoArray {// 双指针public void merge1(int[] nums1, int m, int[] nums2, int n) {int p1 0;int p2 0;int[] sorted new int[m n];int cur 0;int i 0;while (p1 m p2 n) {if (nums1[p1] nums2[p2]) {sorted[i] nums1[p1];} else {sorted[i] nums2[p2];}}while (p1 m) {sorted[i] nums1[p1];}while (p2 n) {sorted[i] nums2[p2];}for (i 0; i m n; i) {nums1[i] sorted[i];}}// 逆向双指针public void merge2(int[] nums1, int m, int[] nums2, int n) {int i m - 1;int j n - 1;int p nums1.length - 1;while (i 0 j 0) {if (nums1[i] nums2[j]) {nums1[p] nums1[i];i--;} else {nums1[p] nums2[j];j--;}p--;}while ( j 0) {nums1[p] nums2[j];j--;p--;}}}
复杂度
时间复杂度O(N)空间复杂度双指针O(N)逆向双指针O(1)
二分查找
int binarySearch(int[] nums, int target) {int left 0, right nums.length - 1; while(left right) {int mid left (right - left) / 2;if (nums[mid] target) {left mid 1;} else if (nums[mid] target) {right mid - 1; } else if(nums[mid] target) {// 直接返回return mid;}}// 直接返回return -1;
}
33. 搜索旋转排序数组⭐️⭐️
思路
代码
public class SearchInRotatedSortedArray {/*nums [4,5,6,7,0,1,2]例如 target 5, 目标值在左半段因此在 [4, 5, 6, 7, inf, inf, inf] 这个有序数组里找就行了例如 target 1, 目标值在右半段因此在 [-inf, -inf, -inf, -inf, 0, 1, 2] 这个有序数组里找就行了*/public int search(int[] nums, int target) {int left 0;int right nums.length - 1;while (left right) {int mid left (right - left) / 2;if (nums[mid] target) {return mid;}if (target nums[0]) {if (nums[mid] nums[0]) {nums[mid] Integer.MAX_VALUE;}} else {if (nums[mid] nums[0]) {nums[mid] Integer.MIN_VALUE;}}if (nums[mid] target) {left mid 1;} else {right mid - 1;}}return -1;}}
复杂度
时间复杂度O(logN)空间复杂度O(1)
69. x 的平方根⭐️⭐️
思路
代码
public class Sqrt {public int mySqrt(int x) {int left 0, right x, res -1;while (left right) {int mid left (right - left) / 2;if ((long) mid * mid x) {res mid;left mid 1;} else {right mid - 1;}}return res;}}
复杂度
时间复杂度O(N)空间复杂度O(1)
240. 搜索二维矩阵 II⭐️
思路 代码
public class SearchMatrix {public boolean searchMatrix(int[][] matrix, int target) {int m matrix.length, n matrix[0].length;int i 0, j n - 1;while (i m j 0) {if (matrix[i][j] target) {return true;}if (matrix[i][j] target) {i; // 需要大一点往下移动} else {j--; // 需要小一点往左移动}}return false;}}
复杂度
时间复杂度O(M N)空间复杂度O(1)
34. 在排序数组中查找元素的第一个和最后一个位置⭐️
思路
代码
public class SearchRange {public int[] searchRange(int[] nums, int target) {return new int[]{leftRange(nums, target), rightRange(nums, target)};}private int leftRange(int[] nums, int target) {int left 0, right nums.length - 1;while (left right) {int mid left (right - left) / 2;if (nums[mid] target) {left mid 1;} else if (nums[mid] target) {right mid - 1;} else if (nums[mid] target) {right mid - 1;}}if (left 0 || left nums.length) {return -1;}return nums[left] target ? left : -1;}private int rightRange(int[] nums, int target) {int left 0, right nums.length - 1;while (left right) {int mid left (right - left) / 2;if (nums[mid] target) {left mid 1;} else if (nums[mid] target) {right mid - 1;} else if (nums[mid] target) {left mid 1;}}if (right 0 || right nums.length) {return -1;}return nums[right] target ? right : -1;}}
复杂度
时间复杂度O(log(N))空间复杂度O(1)
34. 在排序数组中查找元素的第一个和最后一个位置⭐️
思路
代码
class Solution {public int[] searchRange(int[] nums, int target) {return new int[]{leftBound(nums, target), rightBound(nums, target)};}private int leftBound(int[] nums, int target) {int left 0, right nums.length - 1;// 搜索区间为 [left, right]while (left right) {int mid left (right - left) / 2;if (nums[mid] target) {// 搜索区间变为 [mid1, right]left mid 1;} else if (nums[mid] target) {// 搜索区间变为 [left, mid-1]right mid - 1;} else if (nums[mid] target) {// 收缩右侧边界right mid - 1;}}// 检查出界情况if (left nums.length || nums[left] ! target) {return -1;}return left;}private int rightBound(int[] nums, int target) {int left 0, right nums.length - 1;while (left right) {int mid left (right - left) / 2;if (nums[mid] target) {left mid 1;} else if (nums[mid] target) {right mid - 1;} else if (nums[mid] target) {// 这里改成收缩左侧边界即可left mid 1;}}// 这里改为检查 right 越界的情况见下图if (right 0 || nums[right] ! target) {return -1;}return right;}}
复杂度
时间复杂度O(log(N))空间复杂度O(N)
162. 寻找峰值⭐️
思路
代码
public class FindPeakElement {public int findPeakElement(int[] nums) {int left 0, right nums.length - 1;while (left right) {int mid left (right - left) / 2;if (nums[mid] nums[mid 1]) {right mid;} else {left mid 1;}}return left;}}
复杂度
时间复杂度O(logN)空间复杂度O(1)
位运算
136. 只出现一次的数字⭐️
思路
代码
public class SingleNumber {public int singleNumber(int[] nums) {int res 0;for (int n : nums) {res ^ n;}return res;}}
复杂度
时间复杂度O(N)空间复杂度O(1)
