[Swift]LeetCode327. 区间和的个数 | Count of Range Sum

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Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.
Range sum S(i, j) is defined as the sum of the elements in numsbetween indices i and j (i ≤ j), inclusive.

Note:
A naive algorithm of O(n2) is trivial. You MUST do better than that.

Example:

[-2,5,-1]
-2
2
[0,0]
[2,2]
[0,2]
-2, -1, 2

---

给定一个整数数组 nums，返回区间和在 [lower, upper] 之间的个数，包含 lower 和 upper。
区间和 S(i, j) 表示在 nums 中，位置从 i 到 j 的元素之和，包含 i 和 j (i ≤ j)。

说明:
最直观的算法复杂度是 O(n2) ，请在此基础上优化你的算法。

示例:

[-2,5,-1]
-2
2
[0,0]
[2,2]
[0,2]，
-2, -1, 2。

---

140 ms

class Solution {
    func countRangeSum(_ nums: [Int], _ lower: Int, _ upper: Int) -> Int {
        let count = nums.count
        if count == 0 {
            return 0
        }
        var sums = Array(repeating: 0, count: count+1)
        
        for i in 1..<count+1 {
            sums[i] = sums[i-1] + nums[i-1]
        }
        
        let maxSum = sums.max()!
        
        func mergeSort(_ low : Int, _ high : Int) -> Int {
            if low == high {
                return 0
            }
            let mid = (low + high) >> 1
            var res = mergeSort(low, mid) + mergeSort(mid+1, high)
            var x = low, y = low
            for i in mid+1..<high+1 {
                while x <= mid && sums[i] - sums[x] >= lower {
                    x += 1
                }
                while y<=mid && sums[i] - sums[y] > upper {
                    y += 1
                }
                res += (x-y)
            }
            
            let sli = Array(sums[low..<high+1])
            
            var l = low, h = mid + 1
            
            for i in low..<high+1 {
                x = l <= mid ? sli[l - low] : maxSum
                y = h <= high ? sli[h - low] : maxSum
                
                if x < y {
                    l += 1
                }else {
                    h += 1
                }
                sums[i] = min(x,y)
            }
            
            return res
        }
        
        return mergeSort(0, count)
    }
}