leetcode-628-Maximum Product of Three Numbers]
2017-12-25 11:33
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class Solution {
public:
int maximumProduct(vector<int>& nums) {
sort(nums.begin(),nums.end());
int a=nums.size();
return nums[a-3]*nums[a-2]*nums[a-1];
}
};
发现有负的,没辙,
Given an integer array, find three numbers whose product is maximum and output the maximum product.
Example 1:
Example 2:
Note:
The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.
思路:
首先排序,然后分别判断数组元素最大值是正是负情况。
感觉写出来 超级啰嗦 惨不忍睹,于是看到了如下代码,
醍醐灌顶,五体投地。 排序过后,依次讨论前三个,后三个,以及后两个跟第一个,前两个跟最后一个。
发现上面的不行啊;
答案
class Solution {
public:
int maximumProduct(vector<int>& nums) {
int len=nums.size();
sort(nums.begin(),nums.end());//将数组排序
int max1=nums[len-1]*nums[len-2]*nums[len-3];//计算第一种情况的最大值,为最后三个数
int max2=nums[0]*nums[1]*nums[len-1];//计算第二种情况的最大值为前两个数和最后一个数的乘积
return max1>max2?max1:max2;//返回两个中大的那个
}
};
public:
int maximumProduct(vector<int>& nums) {
sort(nums.begin(),nums.end());
int a=nums.size();
return nums[a-3]*nums[a-2]*nums[a-1];
}
};
发现有负的,没辙,
Given an integer array, find three numbers whose product is maximum and output the maximum product.
Example 1:
Input: [1,2,3] Output: 6
Example 2:
Input: [1,2,3,4] Output: 24
Note:
The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.
思路:
首先排序,然后分别判断数组元素最大值是正是负情况。
int maximumProduct(vector<int>& nums) { sort(nums.begin(),nums.end()); int len = nums.size(); int a,b,c; c = nums[len-1]; b = nums[len-2]; a = nums[len-3]; if(a>0)return max(nums[0]*nums[1]*c,a*b*c); else if( a ==0 ) { if(len==3)return 0; if(len>=5)return nums[len-5]*nums[len-4]*c;//l两个负数 else return a*b*c; } else if(a<0) { if(c<0 )return a*b*c; if(c>=0 &&b<0 )return nums[0]*nums[1]*c; if(c>=0 && b>0 &&len>=4)return nums[0]*nums[1]*c; if(c>=0 && b>0 &&len==3)return a*b*c; } return 0; }
感觉写出来 超级啰嗦 惨不忍睹,于是看到了如下代码,
醍醐灌顶,五体投地。 排序过后,依次讨论前三个,后三个,以及后两个跟第一个,前两个跟最后一个。
public int maximumProduct(int[] nums) { Arrays.sort(nums); int n = nums.length; int s = nums[n-1] * nums[n-2] * nums[n-3]; s = Math.max(s, nums[n-1] * nums[n-2] * nums[0]); s = Math.max(s, nums[n-1] * nums[1] * nums[0]); s = Math.max(s, nums[2] * nums[1] * nums[0]); return s; }
发现上面的不行啊;
答案
class Solution {
public:
int maximumProduct(vector<int>& nums) {
int len=nums.size();
sort(nums.begin(),nums.end());//将数组排序
int max1=nums[len-1]*nums[len-2]*nums[len-3];//计算第一种情况的最大值,为最后三个数
int max2=nums[0]*nums[1]*nums[len-1];//计算第二种情况的最大值为前两个数和最后一个数的乘积
return max1>max2?max1:max2;//返回两个中大的那个
}
};
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