leetcode Count of Range Sum

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(_n_2) 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.

题目地址： leetcode Count of Range Sum

题意:

给定一个整数组成的数组，求它的所有子区间和坐落于[lower, upper] 的个数。

比如样例中[-2, 5, -1]中，这三个区间的和在[-2,2]之间 [0, 0], [2, 2], [0, 2]

思路

先来看看最朴素的O(n^2)算法，首先算出和，然后枚举区间范围。

public class Solution {
    public int countRangeSum(int[] nums, int lower, int upper) {
        if(nums.length == 0) return 0;
        long[] sum = new long[nums.length + 1];
        for (int i = 0; i < nums.length; i++)
            sum[i + 1] = sum[i] + nums[i];

        int ans = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j <= nums.length; j++) {
                if(lower  <= sum[j] - sum[i] && sum[j] - sum[i] <= upper)
                    ans++;
            }
        }
        return ans;
    }
}

 

题目要求的效率好于O(n^2)的算法，那么要怎么加速呢？

还记得 Count of Smaller Numbers After Self  么？

那时候，我们用Fenwick树或者线段树，先离散化，然后从右向左扫描，每扫描一个数，对小于它的求和。然后更新.....

这题也差不多，需要找满足条件 lower ≤ sum[j] - sum[i - 1] ≤ upper ，也就是lower + sum[i - 1] ≤ sum[j] ≤ upper + sum[i - 1]

我们同样的求出和，然后离散化，接着从右向左扫描，对每个i 查询满足在[ lower + sum[i - 1], upper + sum[i - 1] ]范围内的个数（用线段树或者Fenwick Tree）

这样复杂度就是O(n log n)

线段树

C++

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;
	}
};

 

Binary indexed tree (Fenwick tree)

关于此树的介绍： Binary indexed tree (Fenwick tree)

注意要加入upper 和 lower -1 两个点

(python 版本比C++简洁太多了^ ^,建议看py版本)

C++

typedef long long LL;
class FenwickTree {
	vector<int> sum_array;
	int n;
	inline int lowbit(int x) {
		return x & -x;
	}

public:
	FenwickTree(int n) :n(n), sum_array(n + 1, 0) {}

	void add(int x, int val) {
		while (x <= n) {
			sum_array[x] += val;
			x += lowbit(x);
		}
	}

	int sum(int x) {
		int res = 0;
		while (x > 0) {
			res += sum_array[x];
			x -= lowbit(x);
		}
		return res;
	}
};

class Solution {
public:
	int countRangeSum(vector<int>& nums, int lower, int upper) {
		if (nums.size() == 0) return 0;
		vector<LL> sum_array (nums.size() * 3,0);
		LL sum = 0;
		for (int i = 0; i < nums.size(); i++) {
			sum += nums[i];
			sum_array[i * 3] = sum;
			sum_array[i * 3 + 1] = sum + lower - 1;
			sum_array[i * 3 + 2] = sum + upper;
		}
		sum_array.push_back(upper);
		sum_array.push_back(lower - 1);
		unordered_map<LL, int> index;
		sort(sum_array.begin(), sum_array.end());
		auto end = unique(sum_array.begin(), sum_array.end());
		auto it = sum_array.begin();
		for (int i = 1; it != end;i++,it++) {
			index[*it] = i;
		}
		FenwickTree tree(index.size());
		int ans = 0;
		for (int i = nums.size() - 1; i >= 0; i--) {
			tree.add(index[sum],1);
			sum -= nums[i];
			ans += tree.sum(index[upper + sum]) - tree.sum(index[lower + sum -1]);
		}
		return ans;
	}
};

 

Python

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 res

class 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 = [upper, lower - 1]
        total = 0
        for num in nums:
            total += num
            sum_array += [total, total + lower - 1, total + upper]

        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[upper + total]) - tree.sum(index[lower + total - 1])
        return ans

 

本文是 leectode 327 Count of Range Sum 的题解，

更多leetcode题解见 https://www.hrwhisper.me/leetcode-algorithm-solution/