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 nums
between indices i
and j
(i
≤ j
),inclusive.
Note:
A naive algorithm of O(n2) is trivial. You MUST do better than that.
Example:
Input: nums =,lower =,upper =,Output: 3 Explanation: The three ranges are :,and their respective sums are: .[-2,5,-1]-22[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:
输入: nums =,输出: 3 解释: 3个区间分别是:,它们表示的和分别为: [-2,2],-2,2。
140 ms
1 class Solution { 2 func countRangeSum(_ nums: [Int],_ lower: Int,_ upper: Int) -> Int { 3 let count = nums.count 4 if count == 0 { 5 return 0 6 } 7 var sums = Array(repeating: 0,count: count+1) 8 9 for i in 1..<count+1 { 10 sums[i] = sums[i-1] + nums[i-1] 11 } 12 13 let maxSum = sums.max()! 14 15 func mergeSort(_ low : Int,_ high : Int) -> Int { 16 if low == high { 17 return 0 18 } 19 let mid = (low + high) >> 1 20 var res = mergeSort(low,mid) + mergeSort(mid+1,high) 21 var x = low,y = low 22 for i in mid+1..<high+1 { 23 while x <= mid && sums[i] - sums[x] >= lower { 24 x += 1 25 } 26 while y<=mid && sums[i] - sums[y] > upper { 27 y += 1 28 } 29 res += (x-y) 30 } 31 32 let sli = Array(sums[low..<high+1]) 33 34 var l = low,h = mid + 1 35 36 for i in low..<high+1 { 37 x = l <= mid ? sli[l - low] : maxSum 38 y = h <= high ? sli[h - low] : maxSum 39 40 if x < y { 41 l += 1 42 }else { 43 h += 1 44 } 45 sums[i] = min(x,y) 46 } 47 48 return res 49 } 50 51 return mergeSort(0,count) 52 } 53 }
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