CPP-回文数-KMP-爬楼梯-最大子段和-牛顿迭代法开方-二分查找法

回文数

通过比较后段的数与前段的数的大小关系实现

class Solution {
public:
    bool isPalindrome(int x) {
        if(x<0 ||  (x>0 && x%10 == 0)){
            return false;
        }
        int sum = 0;
        while (x > sum)
        {
            sum  = sum*10 + x%10;
            x/=10;
        }
        return x==sum || x==sum/10;
    }
};

KMP

class Solution {
public:
    vector<int> getnext(string needle){
        vector<int> next(needle.size(),0);
        next[0] = -1;
        int j=0,k=-1;
        while(j<needle.size()-1){
            if(k==-1||needle[j]==needle[k]){
                next[++j]=++k;
            }
            else{
                k=next[k];
            }
        }
        return next;
    }
    //kmp
    int strStr(string haystack, string needle) {
        vector<int> next =getnext(needle);
        int i=0,j=0,hs=haystack.size(),ns=needle.size();
        while(i<haystack.length()&&j<(int)needle.length()){
            if(j==-1||haystack[i]==needle[j]){
                i++;
                j++;
            }
            else{
                j = next[j];
            }
        }
        if(j==needle.size())return i-j;
        return -1;
    }
};

爬楼梯

经典动态规划
即只能一次爬1或2级台阶，求n级台阶时有几种组合方法

class Solution {
public:
    int climbStairs(int n) {
	//正常的动态规划
        // vector<int> dp(n+1);
        // // dp[0]=0;
        // // dp[1]=1;
        // // dp[2]=2;
        // for(int i=0;i<n+1;i++){
        //     if(i<3) dp[i]=i;
        //     else dp[i] = dp[i-1] + dp[i-2];
        //     cout<<dp[i]<<',';
        // }
	//打表
        int dp[] = {0,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,1134903170,1836311903};
        return dp[n];
    }
};

最大子段和

经典动态规划

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int res=INT_MIN,t=0;
        for(int i=0;i<nums.size();i++){
            t+=nums[i];
            if(t>res)res=t;
            if(t<0)t=0;
        }
        return res;
    }
};

牛顿迭代法开方

class Solution {
public:
    int mySqrt(int x) {
        long x1 = x;
        if(!x1)return 0;
        while(x1*x1 > x){
            x1 = (x1 + x/x1 )/2;
        }
        return x1;

    }
};

二分法寻找插入值位置

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        int res1,lef = 0,rig=nums.size()-1;
        while(lef<=rig){
            res1 = (rig+lef)/2;
            if(nums[res1] == target)return res1;
            else if(nums[res1]>target) rig = res1-1;
            else lef = res1+1;
        }
        return lef;
    }
};