Leetcode ---Maximum Subarray 53

class Solution {
public:
   
    int maxSubArray(vector<int>& nums) {
        int curMax=0;
        int res=INT_MIN;
        for(int num:nums){
            curMax=max(curMax+num,num);
            res=max(res,curMax);
        }
        return res;
    }
};

最大连续字串和。转载https://www.cnblogs.com/grandyang/p/4606334.html。

还有最多连续出现的字串,采用后缀数组的方法时间复杂度是O(nlogn) 也是转载https://blog.csdn.net/ysu108/article/details/7795479

#include <iostream>
using namespace std;
 
int con_sub(char *str, char **ret);
 
int main()
{
        char str[] = "abcabcabcabcabcabbbb";
        char *ret = NULL;
        int time = con_sub(str, &ret);
        printf("%s occuers %d times\n", ret, time);
        return 0;
}
 
int con_sub(char *str, char **ret)
{
        int max_time = 0;//连续出现的最多次数
        int ret_len = 0;//连续出现的字符串的长度
        char *addr = NULL;//连续出现字符串的起始地址
 
        int len = strlen(str);
        char **a = (char **)malloc(sizeof(char *)*len);
        //生成后缀数组
        for(int i=0; i<len; i++)
                a[i] = &str[i];
 
        //重复字符串的长度范围为1到(len+1)/2
        for(int i=1; i<=(len+1)/2; i++)
        {
                //当重复的字符串长度为i的时候,如果是连续出现的,那么第j和第j+i个后缀数组前面为重复的字符串
                for(int j=0; j+i<=len-1; j+=i)
                {
                        int k = j;
                        int temp_time = 1;
                        while(k+i <= len-1 && strncmp(a[k], a[k+i], i) == 0)
                        {
                                temp_time++;
                                k += i;
                        }
                        if(temp_time > max_time)
                        {
                                max_time = temp_time;
                                ret_len = i;
                                addr = a[k];
                        }
                }
        }
        *ret = new char[len+1];
        strncpy(*ret, addr, ret_len);
        return max_time;
}


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转载自blog.csdn.net/xianqianshi3004/article/details/81037567