总的链接 : 

面试经典 150 题 - 学习计划 - 力扣（LeetCode）全球极客挚爱的技术成长平台

 

108 . 将有序数组转换成二叉搜索树 : 

链接 : 

. - 力扣（LeetCode）

思路 : 

选择中间位置左边的数字作为根节点，然后递归调用该方法，遍历顺序采用中序遍历 : 

代码 : (java)

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {TreeNode f(int[] nums,int l,int r){if(l>r) return null ;int mid = l + r >> 1 ;TreeNode root = new TreeNode(nums[mid]) ;root.left = f(nums,l,mid-1);root.right = f(nums,mid+1,r);return root ;}public TreeNode sortedArrayToBST(int[] nums) {int n = nums.length ;TreeNode ans = f(nums,0,n-1);return ans ;}
}

148 . 排序链表 ： 

. - 力扣（LeetCode）

归并排序 : 

class Solution {public ListNode sortList(ListNode head) {return sortList(head, null);}public ListNode sortList(ListNode head, ListNode tail) {if (head == null) {return head;}if (head.next == tail) {head.next = null;return head;}ListNode slow = head, fast = head;while (fast != tail) {slow = slow.next;fast = fast.next;if (fast != tail) {fast = fast.next;}}ListNode mid = slow;ListNode list1 = sortList(head, mid);ListNode list2 = sortList(mid, tail);ListNode sorted = merge(list1, list2);return sorted;}public ListNode merge(ListNode head1, ListNode head2) {ListNode dummyHead = new ListNode(0);ListNode temp = dummyHead, temp1 = head1, temp2 = head2;while (temp1 != null && temp2 != null) {if (temp1.val <= temp2.val) {temp.next = temp1;temp1 = temp1.next;} else {temp.next = temp2;temp2 = temp2.next;}temp = temp.next;}if (temp1 != null) {temp.next = temp1;} else if (temp2 != null) {temp.next = temp2;}return dummyHead.next;}
}

427 . 建立四叉树 

题意 : 矩阵代表四叉树，正方形区域的值全相同的时候，那么这个区域代表一个叶节点，否则不是叶节点，它的叶节点就是四分对应矩阵区域所代表的的子树 ;

用递归+前缀和优化 : 

详细请看代码  : 

class Solution {
public:Node* construct(vector<vector<int>>& grid) {int n = grid.size();vector<vector<int>> sums(n + 1, vector<int>(n + 1));for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) sums[i][j] = sums[i - 1][j] + sums[i][j - 1] - sums[i - 1][j - 1] + grid[i - 1][j - 1];//前缀和function<Node*(int, int, int, int)> dfs = [&](int ax, int ay, int bx, int by) {auto sum = sums[bx][by] - sums[bx][ay] - sums[ax][by] + sums[ax][ay];int mx = (ax + bx) >> 1, my = (ay + by) >> 1; // 找到中间值if(sum == (bx - ax) * (by - ay)) // 判断是否全为 1 return new Node(true, true); // 是一个叶节点 ， 并且结束递归if(sum) return new Node(true, false, dfs(ax, ay, mx, my), dfs(ax, my, mx, by), dfs(mx, ay, bx, my), dfs(mx, my, bx, by));return new Node(false, true);//全0的情况};// dfs(r0,c0,r1,c1)代表处理从r0到r1-1行，从c0到c1-1列的部分return dfs(0, 0, n, n);}
};

23 . 合并k个升序链表 

分治合并 :

先将每两个链表进行合并 ， 完成之后得到k/2个链表，然后继续下去，最后就能够得到一个排好序的合并链表;

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* mergeTwoLists(ListNode *a, ListNode *b) {if ((!a) || (!b)) return a ? a : b;ListNode head, *tail = &head, *aPtr = a, *bPtr = b;while (aPtr && bPtr) {if (aPtr->val < bPtr->val) {tail->next = aPtr; aPtr = aPtr->next;} else {tail->next = bPtr; bPtr = bPtr->next;}tail = tail->next;}tail->next = (aPtr ? aPtr : bPtr);return head.next;}ListNode* mergeKLists(vector<ListNode*>& lists) {ListNode *ans = nullptr;for (size_t i = 0; i < lists.size(); ++i) {ans = mergeTwoLists(ans, lists[i]);}return ans;}
};