comments	difficulty	edit_url	rating	source	tags
true	简单	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1464.Maximum%20Product%20of%20Two%20Elements%20in%20an%20Array/README.md	1121	第 191 场周赛 Q1	数组 排序 堆（优先队列）

1464. 数组中两元素的最大乘积

English Version

题目描述

给你一个整数数组 nums，请你选择数组的两个不同下标 i 和 j，使 (nums[i]-1)*(nums[j]-1) 取得最大值。

请你计算并返回该式的最大值。

 

示例 1：

输入：nums = [3,4,5,2]
输出：12 
解释：如果选择下标 i=1 和 j=2（下标从 0 开始），则可以获得最大值，(nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12 。 

示例 2：

输入：nums = [1,5,4,5]
输出：16
解释：选择下标 i=1 和 j=3（下标从 0 开始），则可以获得最大值 (5-1)*(5-1) = 16 。

示例 3：

输入：nums = [3,7]
输出：12

 

提示：

• 2 <= nums.length <= 500
• 1 <= nums[i] <= 10^3

解法

方法一：暴力枚举

双重循环，枚举所有的下标对，求出 $(nums[i]-1)\times (nums[j]-1)$ 的最大值。其中 $i\ne j$。

时间复杂度 $O(n^2)$。

Python3

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        ans = 0
        for i, a in enumerate(nums):
            for b in nums[i + 1 :]:
                ans = max(ans, (a - 1) * (b - 1))
        return ans

Java

class Solution {
    public int maxProduct(int[] nums) {
        int ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                ans = Math.max(ans, (nums[i] - 1) * (nums[j] - 1));
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int ans = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                ans = max(ans, (nums[i] - 1) * (nums[j] - 1));
            }
        }
        return ans;
    }
};

Go

func maxProduct(nums []int) int {
	ans := 0
	for i, a := range nums {
		for _, b := range nums[i+1:] {
			t := (a - 1) * (b - 1)
			if ans < t {
				ans = t
			}
		}
	}
	return ans
}

TypeScript

function maxProduct(nums: number[]): number {
    const n = nums.length;
    for (let i = 0; i < 2; i++) {
        let maxIdx = i;
        for (let j = i + 1; j < n; j++) {
            if (nums[j] > nums[maxIdx]) {
                maxIdx = j;
            }
        }
        [nums[i], nums[maxIdx]] = [nums[maxIdx], nums[i]];
    }
    return (nums[0] - 1) * (nums[1] - 1);
}

Rust

impl Solution {
    pub fn max_product(nums: Vec<i32>) -> i32 {
        let mut max = 0;
        let mut submax = 0;
        for &num in nums.iter() {
            if num > max {
                submax = max;
                max = num;
            } else if num > submax {
                submax = num;
            }
        }
        (max - 1) * (submax - 1)
    }
}

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function maxProduct($nums) {
        $max = 0;
        $submax = 0;
        for ($i = 0; $i < count($nums); $i++) {
            if ($nums[$i] > $max) {
                $submax = $max;
                $max = $nums[$i];
            } elseif ($nums[$i] > $submax) {
                $submax = $nums[$i];
            }
        }
        return ($max - 1) * ($submax - 1);
    }
}

C

int maxProduct(int* nums, int numsSize) {
    int max = 0;
    int submax = 0;
    for (int i = 0; i < numsSize; i++) {
        int num = nums[i];
        if (num > max) {
            submax = max;
            max = num;
        } else if (num > submax) {
            submax = num;
        }
    }
    return (max - 1) * (submax - 1);
}

方法二：排序

对 $nums$ 进行排序，取最后两个元素，计算乘积 $(nums[n-1]-1)\times (nums[n-2]-1)$ 即可。

时间复杂度 $O(nlogn)$。

Python3

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        nums.sort()
        return (nums[-1] - 1) * (nums[-2] - 1)

Java

class Solution {
    public int maxProduct(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        return (nums[n - 1] - 1) * (nums[n - 2] - 1);
    }
}

C++

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        sort(nums.rbegin(), nums.rend());
        return (nums[0] - 1) * (nums[1] - 1);
    }
};

Go

func maxProduct(nums []int) int {
	sort.Ints(nums)
	n := len(nums)
	return (nums[n-1] - 1) * (nums[n-2] - 1)
}

TypeScript

function maxProduct(nums: number[]): number {
    let max = 0;
    let submax = 0;
    for (const num of nums) {
        if (num > max) {
            submax = max;
            max = num;
        } else if (num > submax) {
            submax = num;
        }
    }
    return (max - 1) * (submax - 1);
}

方法三：一次遍历

遍历 $nums$，维护最大值 $a$ 和次大值 $b$。遍历结束，返回 $(a-1)\times (b-1)$。

Python3

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        a = b = 0
        for v in nums:
            if v > a:
                a, b = v, a
            elif v > b:
                b = v
        return (a - 1) * (b - 1)

Java

class Solution {
    public int maxProduct(int[] nums) {
        int a = 0, b = 0;
        for (int v : nums) {
            if (v > a) {
                b = a;
                a = v;
            } else if (v > b) {
                b = v;
            }
        }
        return (a - 1) * (b - 1);
    }
}

C++

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int a = 0, b = 0;
        for (int v : nums) {
            if (v > a) {
                b = a;
                a = v;
            } else if (v > b) {
                b = v;
            }
        }
        return (a - 1) * (b - 1);
    }
};

Go

func maxProduct(nums []int) int {
	a, b := 0, 0
	for _, v := range nums {
		if v > a {
			b, a = a, v
		} else if v > b {
			b = v
		}
	}
	return (a - 1) * (b - 1)
}