leetcode 327.区间和的个数

链接：https://leetcode-cn.com/problems/count-of-range-sum/

第一种：暴力循环 O(N ^ 3)

public static int test(int[] nums, int lower, int upper) {
	// 纯暴力
	int len = nums.length;
	int count = 0;
	for (int i = 0; i < len; i++) {
		// 计算数组和, 以i为边界
		for (int j = 0; j <= i; j++) {
			// 确定右边界 R
			int sum = 0;
			for (int z = j; z <= i; z++) {
				sum += nums[z];
			}
			if (sum >= lower && sum <= upper) count ++;
		}
	}
	return count;
}

第二种：前缀和优化 O(N^2)

public static int test2(int[] nums, int lower, int upper) {
	// 构造前缀和
	int len = nums.length;
	int[] sums =  new int[len];
	sums[0] = nums[0];
	for (int i = 1; i < len; i++) sums[i] = sums[i - 1] + nums[i];

	int count = 0;
	for (int i = 0; i < len; i++) {
		for (int j = 0; j <= i; j++) {
			int num = -1;
			if (j - 1 < 0) {
				num = sums[i];
			}
			else {
				num = sums[i] - sums[j - 1];
			}
			if (num >= lower && num <= upper) count ++;
		}
	}
	return count;
}

前两种只测试了样例，时间上一定通过不了，在10^5数据下，至少需要O(NlogN)

第三种：前缀和 + 归并排序 O（N * logN）

 

/**
 *
 * @param nums  给定数组nums
 * @param lower  判断是否存在一个子数组在[lower, upper]
 * @param upper
 * @return
 */
public int countRangeSum(int[] nums, int lower, int upper) {
	if (nums == null || nums.length < 2) {
		if (nums != null) {
			if (nums[0] >= lower && nums[0] <= upper) return 1;
		}
		return 0;
	}

	// 计算前缀和
	long[] pres = new long[nums.length];
	pres[0] = nums[0];
	for (int i = 1; i < nums.length; i++) pres[i] = pres[i - 1] + nums[i];

	// 传递前缀数组
	return mergeSort(pres, 0, nums.length - 1, lower, upper);
}

private int mergeSort(long[] nums, int l, int r, int lower, int upper) {
	// 在进行merge的时候，假设当前i， 不会包含[0, i]的情况，所以需要进行特判
	if (l == r) {
		if (nums[l] >= lower && nums[l] <= upper) return 1;
		return 0;
	}

	int mid = l + ((r - l) >> 1);
	return mergeSort(nums, l, mid, lower, upper) + mergeSort(nums, mid + 1, r, lower, upper)
			+ merge(nums, l, mid, r, lower, upper);
}

private int merge(long[] nums, int l, int mid, int r, int lower, int upper) {
	// [windowsL, windowsR)
	int windowsL = l;
	int windowsR = l;
	int count = 0;
	for (int i = mid + 1; i <= r; i ++) {
		long newLower = nums[i] - upper;
		long newUpper = nums[i] - lower;
		while (windowsL <= mid && nums[windowsL] < newLower) windowsL++;
		while (windowsR <= mid && nums[windowsR] <= newUpper) windowsR++;
		count += windowsR - windowsL;
	}

	// 正常merge
	long[] help = new long[r - l + 1];
	int i = 0;
	int l1 = l, l2 = mid + 1;
	while (l1 <= mid && l2 <= r) {
		help[i++] = nums[l1] < nums[l2] ? nums[l1 ++] : nums[l2 ++];
	}
	while (l1 <= mid) {
		help[i ++] = nums[l1 ++];
	}
	while (l2 <= r) {
		help[i ++] = nums[l2 ++];
	}
	for (int j = 0; j < help.length; j++) {
		nums[l + j] = help[j];
	}
	return count;
}

总结

        归并解法：当出现某一位置i，去求解左侧或右侧满足题意情况下可行解个数

类似的问题：归并排序求小和（左侧），归并求解右侧nums[j] * 2 < nums[i]的个数。