[leetcode_4] Median of Two Sorted Arrays

说实话，这个题我第一次没解出来，是看了别人的思路才搞定的，中间在边界问题上还错了n次。惭愧。
先说一下解题思路。
一开始我能想到的思路是：merge sort然后输出array[median]即可。但是很明显这样的时间复杂度高于o(log(m+n))，大概是o(m+n)的复杂度。
时间复杂度的限制是解这个题比较让人郁闷的地方。
ok，再看题目条件，两个已经排好序的数组，时间复杂度是log，这样很容易让人想到什么？二分查找？
但是怎么用二分来解这个题？这儿有一点点数学知识。
具体的推导请看链接：http://blog.chinaunix.net/uid-28490468-id-3576490.html
之前看到也是一篇博客，可是已经找不到了。囧

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ListNode *l;

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        l1 = Reverse(l1);
        l2 = Reverse(l2);
        ListNode *ans = NULL;
        ListNode *ansbackup = NULL;
        int sum = 0;
        int now = 0;
        int next = 0;
        while(true)
        {
            if(l1 == NULL || l2 == NULL)
                break;
            sum = l1->val + l2->val+next;
            now = sum % 10;
            next = sum / 10;
            ListNode *tmp = new ListNode(now);
            if(ans == NULL)
            {
                ans = tmp;
                ansbackup = ans;
            }
            else
            {
                ans->next = tmp;
                ans = ans->next;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
        if(l1 != NULL)
        {
            while(true)
            {
                if(l1 == NULL) break;
                sum = l1->val + next;
                now = sum % 10;
                next = sum / 10;
                ListNode *tmp = new ListNode(now);
                ans->next = tmp;
                ans = ans->next;
                l1 = l1->next;
            }
        }
        if(l2 != NULL)
        {
            while(true)
            {
                if(l2 == NULL) break;
                sum = l2->val + next;
                now = sum % 10;
                next = sum / 10;
                ListNode *tmp = new ListNode(now);
                ans->next = tmp;
                ans = ans->next;
                l2 = l2->next;
            }
        }
        if(next != 0)
        {
            ListNode *tmp = new ListNode(next);
            ans->next = tmp;
            ans = ans->next;
        }
        return ansbackup;
    }
    ListNode *Reverse(ListNode *l)
    {
        ListNode *before;
        ListNode *now;
        ListNode *next;
        if(l->next == NULL)
            return l;
        before = l;
        now = l->next;
        while(now == NULL)
        {
            next = now->next;
            now->next = before;
            before = now;
            now = next;
        }
        return before;
    }
};