leetcode : 327. Count of Range Sum ： 连续和在指定区间内

327. Count of Range Sum

My Submissions

Question

Total Accepted: 3832 Total Submissions: 15268 Difficulty: Hard

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.

这道题目肯定做过，在本科的时候，当时还认真思考过，现在做仍然没做出来，忘记了怎么做，看了题解才有种似曾相识的感觉。

这道题目确实需要技巧~

题意：一组数字，如果连续的数字之和在指定范围内，数量加1，计算总共在范围内的个数。

解法：

1）首先计算sums[]，i表示从0到i的数字之和。问题就转化成寻找(i<j),sums[j]-sums[i]在指定范围内。目前仍然是O（N*N)

2）借助归并排序，合并的过程中，计算满足上面要求的个数。即：i在前一半，j在后一半，且sums[j]-sums[i]在指定范围内，然后在进行合并称有序的sums数组。

因为合并时，前一半和后一般的数量已计算出且数组有序，因此，计算个数和合并的时间复杂都均是O（N）

整体的时间复杂度是O（N*logN）

public class Solution {   public int merge(long[] sums, int lower, int upper, int start, int end) {if (end - start <= 0)return 0;if (end - start == 1) {if (sums[start] >= lower && sums[start] <= upper)return 1;else {return 0;}}int mid = (start + end) / 2;int temp = merge(sums, lower, upper, start, mid)+ merge(sums, lower, upper, mid, end);long[] tttt = new long[(end - start) + 1];int t=mid;int l = mid;int r = mid;int index = 0;for (int i = start; i < mid; i++) {for (; l < end && lower + sums[i] > sums[l]; l++);for (; r < end && upper + sums[i] >= sums[r]; r++);for(;t<end&&sums[t]<sums[i];t++){tttt[index++]=sums[t];}tttt[index++]=sums[i];temp += (r - l);}while (t < end) {tttt[index++] = sums[t++];}System.arraycopy(tttt, 0, sums, start, end-start);return temp;}public int countRangeSum(int[] nums, int lower, int upper) {long[] sums = new long[nums.length];if (nums == null || nums.length == 0)return 0;sums[0] = nums[0];for (int i = 1; i < nums.length; i++)sums[i] = sums[i - 1] + nums[i];return merge(sums, lower, upper, 0, nums.length);}}