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LeetCode327.java
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99 lines (93 loc) · 3.6 KB
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package problems;
import tree.BalancedTree;
/**
* 归并排序
* 设前缀和数组为 preSum，则问题等价于求所有的下标对 (i,j)，满足 preSum[j]−preSum[i]∈[lower,upper]
* 我们先考虑如下的问题：
* 给定两个升序排列的数组 n1, n2, 试找出所有的下标对 (i,j)，满足 n2[j]-n1[i]∈[lower,upper]
* 在已知两个数组均为升序的情况下，这一问题是相对简单的：我们在 n_2 中维护两个指针 l,r。起初，它们都指向 n_2的起始位置。
* 最后计算出符合条件的下标对的数量
*/
public class LeetCode327 {
public int countRangeSum(int[] nums, int lower, int upper) {
long s = 0;
long[] sum = new long[nums.length + 1];
for (int i = 0; i < nums.length; ++i) {
s += nums[i];
sum[i + 1] = s;
}
return countRangeSumRecursive(sum, lower, upper, 0, sum.length - 1);
}
public int countRangeSumRecursive(long[] sum, int lower, int upper, int left, int right) {
if (left == right) {
return 0;
} else {
int mid = (left + right) / 2;
int n1 = countRangeSumRecursive(sum, lower, upper, left, mid);
int n2 = countRangeSumRecursive(sum, lower, upper, mid + 1, right);
int ret = n1 + n2;
// 首先统计下标对的数量
int i = left;
int l = mid + 1;
int r = mid + 1;
while (i <= mid) {
while (l <= right && sum[l] - sum[i] < lower) {
l++;
}
while (r <= right && sum[r] - sum[i] <= upper) {
r++;
}
ret += r - l;
i++;
}
// 随后合并两个排序数组
long[] sorted = new long[right - left + 1];
int p1 = left, p2 = mid + 1;
int p = 0;
while (p1 <= mid || p2 <= right) {
if (p1 > mid) {
sorted[p++] = sum[p2++];
} else if (p2 > right) {
sorted[p++] = sum[p1++];
} else {
if (sum[p1] < sum[p2]) {
sorted[p++] = sum[p1++];
} else {
sorted[p++] = sum[p2++];
}
}
}
for (int j = 0; j < sorted.length; j++) {
sum[left + j] = sorted[j];
}
return ret;
}
}
}
/**
* 平衡二叉搜索树
* 运用到了前缀和二分的思想
* 我们可以从左到右扫描前缀和数组。对于 preSum[j] 而言，
* 首先进行两次查询，得到区间 [preSum[j]−upper,preSum[j]−lower] 内的整数数量；随后再将 preSum[j] 插入到平衡树中。
*/
class LeetCode327_1 {
public int countRangeSum(int[] nums, int lower, int upper) {
long sum = 0;
long[] preSum = new long[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
preSum[i + 1] = sum;
}
BalancedTree tree = new BalancedTree();
int ret = 0;
for (long x : preSum) {
long numLeft = tree.lowerBound(x - upper);
int rankLeft = (numLeft == Long.MAX_VALUE ? (int)(tree.getSize() + 1) : tree.rank(numLeft)[0]);
long numRight = tree.upperBound(x - lower);
int rankRight = (numRight == Long.MAX_VALUE ? (int)(tree.getSize()) : tree.rank(numRight)[0] - 1);
ret += (rankRight - rankLeft + 1);
tree.insert(x);
}
return ret;
}
}