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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 nums
between indices i
and j
(i
≤ j
), inclusive.git
Note:
A naive algorithm of O(n2) is trivial. You MUST do better than that.github
Example:算法
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
給定一個整數數組 nums
,返回區間和在 [lower, upper]
之間的個數,包含 lower
和 upper
。
區間和 S(i, j)
表示在 nums
中,位置從 i
到 j
的元素之和,包含 i
和 j
(i
≤ j
)。數組
說明:
最直觀的算法複雜度是 O(n2) ,請在此基礎上優化你的算法。微信
示例:優化
輸入: nums = , lower = , upper = , 輸出: 3 解釋: 3個區間分別是: , , 它們表示的和分別爲: [-2,5,-1]-22[0,0][2,2][0,2],-2, -1, 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 }