Count of Range Sum(leetcode)

Count of Range Sum

Count of Range Sum
题目

题目解析

解决办法
sum数数

归并排序

题目

leetcode题目

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

.

题目解析

题目的要求就是求原数组的子数组个数，子数组满足“子数组的数之和在给定的范围

[lower, upper]

内”条件。

例子中，

nums

=

[-2, 5, -1]

，子数组为

[-2]

=>

[0, 0]

[5]

=>

[1, 1]

[-1]

=>

[2, 2]

[-2, 5]

=>

[0, 1]

[5, -1]

=>

[1, 2]

[-2, 5, -1]

=>

[0, 2]

子数组的总和在

[-2, 2]

中的子数组为

[-2]

、

[-1]

、

[-2, 5, -1]

，个数为3。

解决办法

1. sum数数

令

sums[i] = nums[0] + nums[1] + … + nums[i]

。

找到满足条件

lower <= sums[j] - sums[i] <= upper

的区间

[i, j](i <= j)

，即我们需要找到满足

sums[j] - upper <= sums[i] <= sums[j] - lower

的

i

的个数。

PS：函数

distance(iterator start, iterator end)

，迭代器start可以通过某种方法到达end，函数distance返回的是start到end的个数。

class Solution {
public:
int countRangeSum(vector<int>& nums, int lower, int upper) {
int res = 0;
long long sum = 0;
multiset<long long> sums;
sums.insert(0);
for (int i = 0; i < nums.size(); ++i) {
sum += nums[i];
res += distance(sums.lower_bound(sum - upper), sums.upper_bound(sum - lower));
sums.insert(sum);
}
return res;
}
};

2. 归并排序

将数组

sums

分为两部分：左半部left和右半部right。由于数组

sums

在构建的时候已经呈递增顺序，使得

left

和

right

已经排好序了。当我们遍历

left

时候，在

right

找到满足以下条件的

j, k

：

sums[j] - sums[i] > upper

，

j

是满足该不等式的第一个下标

sums[k] - sums[i] >= lower

，

k

是满足该不等式的第一个下标

那么在[lower, upper]之间的区间的个数是

j - k

。

同时我们也需要另一个下标t，用来拷贝所有满足sums[t] < sums[i]到一个寄存器Cache中以完成混合排序的过程。

class Solution {
public:
int countRangeSum(vector<int>& nums, int lower, int upper) {
int s = nums.size();
vector<long> sums(s + 1, 0);
for (int i = 0; i < nums.size(); ++i) {
sums[i + 1] = sums[i] + nums[i];
}
return countAndMergeSort(sums, 0, sums.size(), lower, upper);
}
int countAndMergeSort(vector<long> &sums, int start, int end, int lower, int upper) {
if (end - start <= 1) return 0;
int mid = start + (end - start) / 2;
int count = countAndMergeSort(sums, start, mid, lower, upper) + countAndMergeSort(sums, mid, end, lower, upper);
int j = mid;
int k = mid;
int t = mid;
vector<int> cache(end - start, 0);
for (int i = start, r = 0; i < mid; i++, r++) {
while (k < end && sums[k] - sums[i] < lower) k++;
while (j < end && sums[j] - sums[i] <= upper) j++;
while (t < end && sums[t] < sums[i]) cache[r++] = sums[t++];
cache[r] = sums[i];
count += j - k;
}
copy(cache.begin(), cache.begin() + t - start, sums.begin() + start);
return count;
}
};