leetcode题解日练--2016.9.14

平常心

今日题目：

1、最长包含K次重复子串

2、实现平方根sqrt(x)

今日摘录：

说那个话的时候是在那样的一种心醉的情形下，简直什么都可以相信，自己当然绝对相信那不是谎话。
——《半生缘》

395. Longest Substring with At Least K Repeating Characters | Difficulty: Medium

Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input:
s = “aaabb”, k = 3

Output:
3

The longest substring is “aaa”, as ‘a’ is repeated 3 times.
Example 2:

Input:
s = “ababbc”, k = 2

Output:
5

The longest substring is “ababb”, as ‘a’ is repeated 2 times and ‘b’ is repeated 3 times.

tag：
题意：找到最长子串，需要满足这个子串中每个字符出现次数均要大于K次。

思路：
1、如何找到这样的子串，找到一头一尾(first、last)，并且中间的都满足出现次数大于等于K。每找到一个这样的last就更新一次，这样从一个first找到最后之后，每次查找之后将first移动到最后一个找到符合的last字符的后面一位。

class Solution {public:    int longestSubstring(string s, int k) {        int max_len = 0;        int first,last;        int n = s.size();        for(first = 0;first+k<=n;)        {            int max_last=first;            int mask = 0;            int count[26] = {0};            for(last = first;last<n;last++)            {                int i = s[last]-'a';                count[i]++;                if(count[i]<k)  mask |=(1<<i);                else    mask&=(~(1<<i));                if(mask==0)                 {                    max_len = max(max_len,last-first+1);                    max_last = last;                }            }            if(max_len>=n-first) break;            first = max_last+1;        }        return max_len;    }};

结果：3ms

2、递归写法，每次从当前区间[first,last]中去查找区间，先遍历一次做一次统计，然后统计完之后去找一个在[first,last]中出现大于等于K次的字符组成的区间，再递归去判断这个更小的区间。递归终止条件是更小的区间就是原区间，返回区间长度。

class Solution {public:    int helper(const string& s,int k,int first,int last)    {        if(first>last)  return 0;        int cnt[26] = {0};        for(int i=first;i<last;i++)        {            cnt[s[i]-'a']++;        }         int max_len=0;        int l=0,r=0;        for(l = first;l<last;)        {            while(l<last && cnt[s[l]-'a'] <k) l++;            if(l==last) break;            r = l;            while(r<last && cnt[s[r]-'a']>=k) r++;            if(l==first && r==last) return r-l;            max_len = max(max_len,helper(s,k,l,r));            l = r;        }        return max_len;    }    int longestSubstring(string s, int k) {        int n = s.size();        if(n==0)    return 0;        return helper(s,k,0,n);    }};

结果：3ms

69. Sqrt(x) | Difficulty: Medium

Implement int sqrt(int x).

Compute and return the square root of x.

tag：数学|二分

题意：实现平方根函数
思路：
1、用牛顿法去做，
x_{{n+1}}=x_{n}-{\frac {f(x_{n})}{f’(x_{n})}}

class Solution {public:    int mySqrt(int x) {        long long  r = x;        while(r*r>x)        {            r = (r+x/r)/2;        }        return r;    }};

结果：6ms

2、二分查找
相当于找第一个<=sqrt(x)的整数，这种情况下要使用偏右的二分查找方法

class Solution {public:    int mySqrt(int x) {        if(x==0)    return 0;        int l = 1,r = x;        while(l<r)        {            int mid = l+(r-l+1)/2;            if(mid>x/mid)   r = mid-1;            else if(mid<x/mid) l=mid;            else    return mid;        }        return l;    }};

结果：6ms