Forty-eight daily algorithms: plus one (an array of decimal digits plus one digit) and SQRT (X)

Given the decimal number represented by an array, add one. The result is still represented in a decimal array. Here, we mainly note that the highest bit (digit [0]) still has an incoming bit, that is, overflow.

Given a non-negative number represented as an array of digits, plus one to the number.

The digits are stored such that the most significant digit is at the head of the list.

<span style="font-size:18px;">class Solution {public:    vector<int> plusOne(vector<int> &digits) {        int len = digits.size();        if(len == 0)            return digits;        int carry = 1;int temp = 0;        for(int i = len-1; i>=0; i--)        {            temp = (digits[i] + carry)/10;            digits[i] = (digits[i] + carry)%10;            carry = temp;        }        if(carry > 0)        {            vector<int> ret(len+1, 0);            ret[0] = 1;            return ret;        }        else            return digits;    }};</span>

SQRT (x ):

Class solution {public: int SQRT (int x) {double diff = 0.01; // error int low = 0; int high = x; while (low <= high) {// pay attention to cross-border! So double mid = low + (high-low)/2; If (ABS (mid * mid-x) <= diff) {return (INT) Mid ;} else if (x> mid * Mid + diff) {LOW = (INT) Mid + 1;} else if (x <mid * mid-diff) {high = (INT) mid-1 ;}/// when it cannot be found, this is low, the high pointer has been crossed, take a smaller value, that is, high return high ;}};

Forty-eight daily algorithms: plus one (an array of decimal digits plus one digit) and SQRT (X)