989.Add to Array-Form of Integer
989.Add to Array-Form of Integer
难度:Easy
For a non-negative integer X, the array-form of X is an array of its digits in left to right order. For example, if X = 1231, then the array form is [1,2,3,1].
Given the array-form A of a non-negative integer X, return the array-form of the integer X+K.
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Example 1:
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Input: A = [1,2,0,0], K = 34
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Output: [1,2,3,4]
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Explanation: 1200 + 34 = 1234
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Example 2:
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Input: A = [2,7,4], K = 181
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Output: [4,5,5]
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Explanation: 274 + 181 = 455
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Example 3:
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Input: A = [2,1,5], K = 806
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Output: [1,0,2,1]
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Explanation: 215 + 806 = 1021
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Example 4:
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Input: A = [9,9,9,9,9,9,9,9,9,9], K = 1
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Output: [1,0,0,0,0,0,0,0,0,0,0]
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Explanation: 9999999999 + 1 = 10000000000
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Note:
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1 <= A.length <= 10000
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0 <= A[i] <= 9
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0 <= K <= 10000
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If A.length > 1, then A[0] != 0
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可以全部轉化成string,然後計算加法,這樣不會溢出,不過運行時間較長。另一種方法是,對K每加一位,再向前更新。
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class Solution {
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public:
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vector<int> addToArrayForm(vector<int>& A, int K) {
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if(K==0) return A;
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bool flag=false;
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for(int i=A.size()-1;i>=0;i--)
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{
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if(K){
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K+=A[i];
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A[i]=K%10;
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K/=10;
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}
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else
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{
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flag=true;
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break;
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}
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}
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// cout<<flag<<endl;
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if(flag || K==0) return A;
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int len=log10(K)+1;
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vector<int >res(len+A.size(),0);
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for(int i=len-1;i>=0;i--)
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{
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res[i]=K%10;
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K/=10;
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}
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for(int i=len;i<res.size();i++)
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res[i]=A[i-len];
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return res;
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}
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};
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從個位算起,與K相加,將和的個位留在該位置,然後和除以10,繼續向前移位相加。 其中,flag表示在計算求和過程中,K如果爲0,則不必向高位繼續計算,直接返回A矩陣。 然後計算遍歷完A之後K的大小,將其轉化爲矩陣並與A合併得最終結果。
执行用时 :164 ms, 在所有 C++ 提交中击败了91.72%的用户 内存消耗 :12.4 MB, 在所有 C++ 提交中击败了84.39%的用户
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