2219. Maximum Sum Score of Array

English Version

题目描述

You are given a 0-indexed integer array nums of length n.

The sum score of nums at an index i where 0 <= i < n is the maximum of:

• The sum of the first i + 1 elements of nums.
• The sum of the last n - i elements of nums.

Return the maximum sum score of nums at any index.

 

Example 1:

Input: nums = [4,3,-2,5]
Output: 10
Explanation:
The sum score at index 0 is max(4, 4 + 3 + -2 + 5) = max(4, 10) = 10.
The sum score at index 1 is max(4 + 3, 3 + -2 + 5) = max(7, 6) = 7.
The sum score at index 2 is max(4 + 3 + -2, -2 + 5) = max(5, 3) = 5.
The sum score at index 3 is max(4 + 3 + -2 + 5, 5) = max(10, 5) = 10.
The maximum sum score of nums is 10.

Example 2:

Input: nums = [-3,-5]
Output: -3
Explanation:
The sum score at index 0 is max(-3, -3 + -5) = max(-3, -8) = -3.
The sum score at index 1 is max(-3 + -5, -5) = max(-8, -5) = -5.
The maximum sum score of nums is -3.

 

Constraints:

• n == nums.length
• 1 <= n <= 105
• -105 <= nums[i] <= 105

解法

前缀和。

Python3

class Solution:
    def maximumSumScore(self, nums: List[int]) -> int:
        s = [0] + list(accumulate(nums))
        return max(max(s[i + 1], s[-1] - s[i]) for i in range(len(nums)))

Java

class Solution {
    public long maximumSumScore(int[] nums) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        long ans = Long.MIN_VALUE;
        for (int i = 0; i < n; ++i) {
            ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
        }
        return ans;
    }
}

TypeScript

function maximumSumScore(nums: number[]): number {
    const n = nums.length;
    let s = new Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    let ans = -Infinity;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
    }
    return ans;
}

C++

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

Go

func maximumSumScore(nums []int) int64 {
	n := len(nums)
	s := make([]int64, n+1)
	for i, v := range nums {
		s[i+1] = s[i] + int64(v)
	}
	var ans int64 = math.MinInt64
	for i := 0; i < n; i++ {
		ans = max(ans, max(s[i+1], s[n]-s[i]))
	}
	return ans
}

func max(a, b int64) int64 {
	if a > b {
		return a
	}
	return b
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var maximumSumScore = function (nums) {
    const n = nums.length;
    let s = new Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    let ans = -Infinity;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
    }
    return ans;
};

...