leetcode Count of Range Sum

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:
Given nums = [-2, 5, -1], lower = -2, upper = 2,
Return 3.

The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

这里只贴代码，解释什么的、后续更新 见原文 http://www.hrwhisper.me/leetcode-count-of-range-sum/

typedef long long LL;struct SegmentTreeNode {LL L, R;int cnt;SegmentTreeNode *left, *right;SegmentTreeNode(LL L, LL R) :L(L), R(R), cnt(0), left(NULL), right(NULL) {}};class SegmentTree {SegmentTreeNode * root;SegmentTreeNode * buildTree(vector<LL> &nums, int L, int R) {if (L > R) return NULL;SegmentTreeNode * root = new SegmentTreeNode(nums[L], nums[R]);if (L == R) return root;int mid = (L + R) >> 1;root->left = buildTree(nums, L, mid);root->right = buildTree(nums, mid + 1, R);return root;}void update(SegmentTreeNode * root, LL val) {if (root && root->L <= val &&  val <= root->R) {root->cnt++;update(root->left, val);update(root->right, val);}}int sum(SegmentTreeNode * root, LL L, LL R) {if (!root || root->R < L ||  R < root->L ) return 0;if (L <= root->L  && root->R <= R) return root->cnt;return sum(root->left, L, R) + sum(root->right, L, R);}public:SegmentTree(vector<LL> &nums, int L, int R) { root = buildTree(nums, L, R); }int sum(LL L, LL R) {return sum(root, L, R);}void update(LL val) {update(root, val);}};class Solution {public:int countRangeSum(vector<int>& nums, int lower, int upper) {if (nums.size() == 0) return 0;vector<LL> sum_array (nums.size(),0);sum_array[0] = nums[0];for (int i = 1; i < sum_array.size(); i++) {sum_array[i] = nums[i] + sum_array[i - 1];}LL sum = sum_array[sum_array.size() - 1];sort(sum_array.begin(), sum_array.end());auto t = unique(sum_array.begin(), sum_array.end());SegmentTree tree(sum_array, 0, t - sum_array.begin() - 1);int ans = 0;for (int i = nums.size() - 1; i >= 0; i--) {tree.update(sum);sum -= nums[i];ans += tree.sum(lower + sum,upper + sum);}return ans;}};

class FenwickTree(object):    def __init__(self, n):        self.sum_array = [0] * (n + 1)        self.n = n    def lowbit(self, x):        return x & -x    def add(self, x, val):        while x <= self.n:            self.sum_array[x] += val            x += self.lowbit(x)    def sum(self, x):        res = 0        while x > 0:            res += self.sum_array[x]            x -= self.lowbit(x)        return resclass Solution(object):    def countRangeSum(self, nums, lower, upper):        """        :type nums: List[int]        :type lower: int        :type upper: int        :rtype: int        """        if not nums: return 0        sum_array = []        total = 0        for num in nums:            total += num            sum_array += [total] + [total + lower - 1] + [total + upper]        sum_array += [upper] + [lower - 1]        index = {}        for i, x in enumerate(sorted(set(sum_array))):            index[x] = i + 1         tree = FenwickTree(len(index))        ans = 0        for i in xrange(len(nums) - 1, -1, -1):            tree.add(index[total], 1)            total -= nums[i]            ans += tree.sum(index.get(upper + total, 0)) - tree.sum(index.get(lower + total - 1, 0))        return ans

本文是 leectode 327 Count of Range Sum 的题解，

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