缺失的正整数

/*** 0～n-1中缺失的数字*/
public class MissingNum {public int missingNumber(int[] nums) {int i = 0;int j = nums.length - 1;while (i < j) {int m = i + ((j - i) >> 2);if (nums[m] == m) i = m + 1;else j = m - 1;}return nums[i] == i ? nums[i] + 1 : nums[i] - 1;}public static void main(String[] args) {int[] arr = {0, 1, 2, 3, 4, 5, 6, 7, 9};MissingNum missingNum = new MissingNum();System.out.println(missingNum.missingNumber(arr));}
}

数组中重复的数字

 //leetcode 指 Offer 03. 数组中重复的数字public int findRepeatNumber(int[] nums) {HashMap<Integer, Integer> map = new HashMap<>();int res = nums[0];for (int num : nums) {if (map.get(num) == null) {map.put(num, 1);} else {res = num;break;}}return res;}

盛水最多的容器

 //leetcode-11 盛最多水的容器public int maxArea(int[] height) {int left = 0;int right = height.length - 1;int maxArea = 0;/*  while (left < right) {int width = right - left;int deep = Math.min(height[left], height[right]);maxArea = Math.max(maxArea, width*deep);if (height[left] < height[right]) {left++;} else {right--;}}return maxArea;*/while (left < right) {maxArea = height[left] < height[right] ?Math.max(maxArea, (right - left) * height[left++]) :Math.max(maxArea, (right - left) * height[right--]);}return maxArea;}

 

二分查找

//二分查找：初始值left=0,right=length-1,搜索闭区间[left,right]; while循环条件left<=right;public static int binarySearch(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;} else if (nums[mid] < target) {left = mid + 1;} else if (nums[mid] > target) {right = mid - 1;}}//return left;return nums[left] == target ? left : -1;}//左边界查找：public static int leftBoundarySearch(int[] nums, int target) {int left = 0;int right = nums.length;while (left < right) {int mid = left + (right - left) / 2;if (nums[mid] == target) {right = mid;} else if (nums[mid] < target) {left = mid + 1;} else if (nums[mid] > target) {right = mid;}}//return left;if (nums.length == left) {return -1;}return nums[left] == target ? left : -1;}//右边界查找：public static int rightBoundarySearch(int[] nums, int target) {int left = 0;int right = nums.length;while (left < right) {int mid = left + (right - left) / 2;if (nums[mid] == target) {left = mid + 1;} else if (nums[mid] < target) {left = mid + 1;} else if (nums[mid] > target) {right = mid;}}//return left - 1;if (left == 0) {return -1;}return nums[left - 1] == target ? left - 1 : -1;}

合并两个有序链表

//合并两个有序链表public static ListNode mergeTwoLists1(ListNode l1, ListNode l2) {ListNode res = new ListNode(0);ListNode head = res;//指向头结点的引用while (l1 != null && l2 != null) {if (l1.value < l2.value) {res.next = l1;l1 = l1.next;} else {res.next = l2;l2 = l2.next;}res = res.next;}if (l2 == null) {res.next = l1;} else {res.next = l2;}return head.next;}

找到两个单链表相交的起始结点(判断链表是否相交)

 //找到两个单链表相交的起始结点(判断链表是否相交)//走到尽头见不到你，于是走过你来时的路，等到相遇时才发现，你也走过我来时的路//我先走我的路，再走你的路，你先走你的路，再走我的路，这样咱俩走的路程就一样了，速度一样，那么肯定在咱俩两条路的交叉口相遇，并一起走向终点。这思路，给跪了！public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {if (headA == null || headB == null) {return null;}ListNode A = headA;ListNode B = headB;// 直到两人相遇while (A != B) {if (A != null) {// 如果A没走到尽头就一直走下去A = A.next;} else {// 直到尽头也没遇见B，所以去往B走过的路A = headB;}if (B != null) {B = B.next;} else {B = headA;}}// 返回A\B第一次相遇的地方return B;}

求单链表是否有环（快、慢指针（龟兔赛跑思想,如果出现环形，那么必然相遇），快指针每次移动2步，慢指针每次移动一步

 //求单链表是否有环（快、慢指针（龟兔赛跑思想,如果出现环形，那么必然相遇），快指针每次移动2步，慢指针每次移动一步public static boolean hasCycle(ListNode head) {if(head == null || head.next == null){return false;}ListNode slow = head;ListNode fast = head.next;while (slow != fast){if (fast == null||fast.next == null){return false;}slow = slow.next;fast =fast.next.next;}//哈希表方案：每次遍历到一个节点时，判断该节点此前是否被访问过。
/*        Set<ListNode> seen = new HashSet<>();while (head!=null){if (!seen.add(head)){return true;}head = head.next;}return false;*/return true;}