(LeetCode OJ) Plus One [66]

Source: Internet
Author: User

(LeetCode OJ) Plus One [66]

66. Plus One

My Submissions QuestionTotal Accepted: 77253 Total Submissions: 242942 Difficulty: Easy

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.

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// First clear the train of thought: the question is that an array is used to represent a non-negative integer, and each bit of the number is stored in the array. // it is clear that the end of the array is the second bit of the number, the start is the highest bit // if the low position is less than 9, it is obvious that it can be directly + 1 // if it is = 9, then the position is zero and the last two digits + 1, similarly, if the value is smaller than 9, it is directly + 1 if it is smaller than 9. Otherwise, it is carried to the front of the array. // an extreme case: for example, 999, after the carry is 1000, the class Solution {public: vector is extended.
PlusOne (vector
& Digits) {vector
Ans = digits; int I = 0; for (I = ans. size ()-1; I> = 0; I --) // The traversal starts from the low position, and the reverse traversal {if (I = 0 & ans [0] = 9) {// processing extreme cases ans [I] = 0; vector
: Iterator ansiter = ans. begin (); ans. insert (ansiter, 1); break;} if (ans [I]! = 9) {ans [I] + = 1; break;} else {ans [I] = 0 ;}} return ans ;}};

Dirty performance ...............................

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