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中等
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第 50 场双周赛 Q3
位运算
数组
前缀和

English Version

题目描述

给你一个 有序 数组 nums ,它由 n 个非负整数组成,同时给你一个整数 maximumBit 。你需要执行以下查询 n 次:

  1. 找到一个非负整数 k < 2maximumBit ,使得 nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k 的结果 最大化 。k 是第 i 个查询的答案。
  2. 从当前数组 nums 删除 最后 一个元素。

请你返回一个数组 answer ,其中 answer[i]是第 i 个查询的结果。

 

示例 1:

输入:nums = [0,1,1,3], maximumBit = 2
输出:[0,3,2,3]
解释:查询的答案如下:
第一个查询:nums = [0,1,1,3],k = 0,因为 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3 。
第二个查询:nums = [0,1,1],k = 3,因为 0 XOR 1 XOR 1 XOR 3 = 3 。
第三个查询:nums = [0,1],k = 2,因为 0 XOR 1 XOR 2 = 3 。
第四个查询:nums = [0],k = 3,因为 0 XOR 3 = 3 。

示例 2:

输入:nums = [2,3,4,7], maximumBit = 3
输出:[5,2,6,5]
解释:查询的答案如下:
第一个查询:nums = [2,3,4,7],k = 5,因为 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7。
第二个查询:nums = [2,3,4],k = 2,因为 2 XOR 3 XOR 4 XOR 2 = 7 。
第三个查询:nums = [2,3],k = 6,因为 2 XOR 3 XOR 6 = 7 。
第四个查询:nums = [2],k = 5,因为 2 XOR 5 = 7 。

示例 3:

输入:nums = [0,1,2,2,5,7], maximumBit = 3
输出:[4,3,6,4,6,7]

 

提示:

  • nums.length == n
  • 1 <= n <= 105
  • 1 <= maximumBit <= 20
  • 0 <= nums[i] < 2maximumBit
  • nums​​​ 中的数字已经按 升序 排好序。

解法

方法一:位运算 + 枚举

我们先预处理出数组 nums 的异或和 $xs$,即 $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n-1]$

接下来,我们从后往前枚举数组 nums 中的每个元素 $x$,当前的异或和为 $xs$,我们需要找到一个数 $k$,使得 $xs \oplus k$ 的值尽可能大,并且 $k \lt 2^{maximumBit}$

也即是说,我们从 $xs$ 的第 $maximumBit - 1$ 位开始,往低位枚举,如果 $xs$ 的某一位为 $0$,那么我们就将 $k$ 的对应位设置为 $1$,否则我们就将 $k$ 的对应位设置为 $0$。这样,最终得到的 $k$ 就是每一次查询的答案。然后,我们将 $xs$ 更新为 $xs \oplus x$,继续枚举下一个元素。

时间复杂度 $O(n \times m)$,其中 $n$$m$ 分别是数组 numsmaximumBit 的值。忽略答案数组的空间消耗,空间复杂度 $O(1)$

Python3

class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        ans = []
        xs = reduce(xor, nums)
        for x in nums[::-1]:
            k = 0
            for i in range(maximumBit - 1, -1, -1):
                if (xs >> i & 1) == 0:
                    k |= 1 << i
            ans.append(k)
            xs ^= x
        return ans

Java

class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int n = nums.length;
        int xs = 0;
        for (int x : nums) {
            xs ^= x;
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = 0;
            for (int j = maximumBit - 1; j >= 0; --j) {
                if (((xs >> j) & 1) == 0) {
                    k |= 1 << j;
                }
            }
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
        int xs = 0;
        for (int& x : nums) {
            xs ^= x;
        }
        int n = nums.size();
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = 0;
            for (int j = maximumBit - 1; ~j; --j) {
                if ((xs >> j & 1) == 0) {
                    k |= 1 << j;
                }
            }
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
};

Go

func getMaximumXor(nums []int, maximumBit int) (ans []int) {
	xs := 0
	for _, x := range nums {
		xs ^= x
	}
	for i := range nums {
		x := nums[len(nums)-i-1]
		k := 0
		for j := maximumBit - 1; j >= 0; j-- {
			if xs>>j&1 == 0 {
				k |= 1 << j
			}
		}
		ans = append(ans, k)
		xs ^= x
	}
	return
}

TypeScript

function getMaximumXor(nums: number[], maximumBit: number): number[] {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = 0;
        for (let j = maximumBit - 1; j >= 0; --j) {
            if (((xs >> j) & 1) == 0) {
                k |= 1 << j;
            }
        }
        ans[i] = k;
        xs ^= x;
    }
    return ans;
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} maximumBit
 * @return {number[]}
 */
var getMaximumXor = function (nums, maximumBit) {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = 0;
        for (let j = maximumBit - 1; j >= 0; --j) {
            if (((xs >> j) & 1) == 0) {
                k |= 1 << j;
            }
        }
        ans[i] = k;
        xs ^= x;
    }
    return ans;
};

C#

public class Solution {
    public int[] GetMaximumXor(int[] nums, int maximumBit) {
        int xs = 0;
        foreach (int x in nums) {
            xs ^= x;
        }
        int n = nums.Length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = 0;
            for (int j = maximumBit - 1; j >= 0; --j) {
                if ((xs >> j & 1) == 0) {
                    k |= 1 << j;
                }
            }
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}

方法二:枚举优化

与方法一类似,我们先预处理出数组 nums 的异或和 $xs$,即 $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n-1]$

接下来,我们算出 $2^{maximumBit} - 1$,即 $2^{maximumBit}$ 减去 $1$,记为 $mask$。然后,我们从后往前枚举数组 nums 中的每个元素 $x$,当前的异或和为 $xs$,那么 $k=xs \oplus mask$ 就是每一次查询的答案。然后,我们将 $xs$ 更新为 $xs \oplus x$,继续枚举下一个元素。

时间复杂度 $O(n)$,其中 $n$ 是数组 nums 的长度。忽略答案数组的空间消耗,空间复杂度 $O(1)$

Python3

class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        ans = []
        xs = reduce(xor, nums)
        mask = (1 << maximumBit) - 1
        for x in nums[::-1]:
            k = xs ^ mask
            ans.append(k)
            xs ^= x
        return ans

Java

class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int xs = 0;
        for (int x : nums) {
            xs ^= x;
        }
        int mask = (1 << maximumBit) - 1;
        int n = nums.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = xs ^ mask;
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
        int xs = 0;
        for (int& x : nums) {
            xs ^= x;
        }
        int mask = (1 << maximumBit) - 1;
        int n = nums.size();
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = xs ^ mask;
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
};

Go

func getMaximumXor(nums []int, maximumBit int) (ans []int) {
	xs := 0
	for _, x := range nums {
		xs ^= x
	}
	mask := (1 << maximumBit) - 1
	for i := range nums {
		x := nums[len(nums)-i-1]
		k := xs ^ mask
		ans = append(ans, k)
		xs ^= x
	}
	return
}

TypeScript

function getMaximumXor(nums: number[], maximumBit: number): number[] {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const mask = (1 << maximumBit) - 1;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = xs ^ mask;
        ans[i] = k;
        xs ^= x;
    }
    return ans;
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} maximumBit
 * @return {number[]}
 */
var getMaximumXor = function (nums, maximumBit) {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const mask = (1 << maximumBit) - 1;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = xs ^ mask;
        ans[i] = k;
        xs ^= x;
    }
    return ans;
};

C#

public class Solution {
    public int[] GetMaximumXor(int[] nums, int maximumBit) {
        int xs = 0;
        foreach (int x in nums) {
            xs ^= x;
        }
        int mask = (1 << maximumBit) - 1;
        int n = nums.Length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = xs ^ mask;
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}