Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
其实这道题最简单的方法恰恰就是他禁止使用的除法,不光思路简单,操作也十分简单。除去这个方法,就只能使用两个数组,分别计算正序乘积和倒序乘积了。
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int len = nums.size(); //获取数组长度
if(0 == len || 1 == len){
vector<int> ret = nums;
return ret;
}
vector<int> ins(len, nums[0]); //开辟保存正序乘积数组
vector<int> ret(len, nums[len-1]); //开辟保存逆序乘积数组
int mul1 = nums[0], mul2 = nums[len-1];
for(int i = 1; i < len; ++i){
mul1 *= nums[i];
mul2 *= nums[len-i-1];
ins[i] = mul1;
ret[len-i-1] = mul2;
} //计算数组值
ret[0] = ret[1];
for(int i = 1; i < len-1; ++i)
ret[i] = ins[i-1] * ret[i + 1];
ret[len-1] = ins[len-2];
return ret;
}
};
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