7.17 sorting ~findMissing桶排序

to do

1] Merge Sorted Array

注意comment out的一句可省

    void merge(vector& nums1, int m, vector& nums2, int n) {
        int write = m+n-1; // pos to write the next one
        int read1 = m-1;
        int read2 = n-1;
        while (read1!=-1 && read2!=-1) {
            if (nums1[read1]>nums2[read2]) {
                nums1[write--] = nums1[read1--];
            } else {
                nums1[write--] = nums2[read2--];
            }
        }
        // while (read1!=-1) nums1[write--] = nums1[read1--];
        while (read2!=-1) nums1[write--] = nums2[read2--];
    }

2] Merge Two Sorted Lists

    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (!l1 || !l2) return l1? l1 : l2;
        
        ListNode dummy = ListNode(-1);
        ListNode* curr = &dummy;
        for (; l1&&l2; curr=curr->next) {
            if (l1->val < l2->val) {
                curr->next = l1;
                l1 = l1->next;
            } else {
                curr->next = l2;
                l2 = l2->next;
            }
        }
        if (l1 || l2) curr->next = l1? l1 : l2;
        return dummy.next;
    }

2] see below

3] Merge k Sorted Lists

    ListNode* mergeTwo (ListNode* l1, ListNode* l2) {
        if (!l1 || !l2) return l1? l1 : l2;
        
        ListNode dummy = ListNode(-1);
        ListNode* curr=&dummy;
        for (; l1&&l2; curr=curr->next) {
            if (l1->val < l2->val) {
                curr->next = l1;
                l1 = l1->next;
            } else {
                curr->next = l2;
                l2 = l2->next;
            }
        }
        curr->next = l1? l1 : l2;
        return dummy.next;
    }
    
    ListNode* mergeKLists(vector& lists) {
        if (lists.empty()) return nullptr;
        ListNode* ret = nullptr;
        for (int i=0; i

3] Merge k Sorted Lists

学会分析下bigO。。另外mergeTwo的居然是recursion的更快，recursion stack的时空间分析回头来

1. iterate through all lists in search for next-to-insert
2. merge each list with ret
3. partition then merge：

O(nlogk): split takes O(logk), merge takes O(n)??

    ListNode* mergeTwo (ListNode* l1, ListNode* l2) {
        if (!l1 || !l2) return l1? l1 : l2;
        if (l1->val < l2->val) {
            l1->next = mergeTwo(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwo(l1, l2->next);
            return l2;
        }
        return nullptr;
    }
    
    ListNode* splitNmergeR(vector& lists, int l, int r) {
        if (l>r) return nullptr;
        if (l==r) return lists[l];
        if (l+1==r) return mergeTwo(lists[l], lists[r]);
        
        int mid = (l+r)/2;
        return mergeTwo(splitNmergeR(lists, l, mid-1), splitNmergeR(lists, mid, r));
    }
    
    ListNode* mergeKLists(vector& lists) {
        return splitNmergeR(lists, 0, lists.size()-1);
    }

4] Insertion Sort List

mark redo X( 流程图

    ListNode* insertionSortList(ListNode* head) {
        if (!head) return nullptr;
        
        ListNode dummy = ListNode (-1);
        dummy.next = head;
        ListNode* tail = head;
        for (ListNode* curr=head->next, *nextCurr; curr; curr=nextCurr) {
            nextCurr=curr->next;
            tail->next = nullptr;   
            bool inserted = false;
            for (ListNode* prev=&dummy; prev!=tail; prev=prev->next) {
                if (prev->next->val >= curr->val) {
                    curr->next = prev->next;
                    prev->next = curr;
                    inserted = true;
                    break;
                }
            }
            if (!inserted) {
                curr->next = nullptr;
                tail->next = curr;
                tail = curr;
            }
        }
        
        return dummy.next;
    }

5] Sort List

beat 95% hhh, 把for loop body变为afterthought就跌成6%了有意思

    ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
        if (!l1 || !l2) return l1? l1: l2;
        if (l1->val < l2->val) {
            l1->next = mergeTwo(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwo(l1, l2->next);
            return l2;
        }
        return nullptr;
    }
    
    ListNode* sortList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode* slow=head;
        for (ListNode* fast=head; fast->next && fast->next->next; ) {
            fast = fast->next->next;
            slow = slow->next;
        }
        ListNode* l2 = slow->next;
        slow->next = nullptr;
        return mergeTwo(sortList(head), sortList(l2));
    }

7] Sort Colors

double ptrs

    void sortColors(vector& nums) {
        int red = 0, blue = nums.size()-1;
        for (int j=0; j<=blue; ) {
            if (nums[j]==0) {
                nums[red++] = nums[j++];
            } else if (nums[j]==2) {
                swap(nums[blue--], nums[j]);
            } else {
                ++j;
            }
        }
        for (int i=red; i<=blue; ++i) nums[i]=1;
    }