Count Of Range Sum

Count Of Range Sum

题目: 求数组arr的子数组累加和在range [low, high]范围（闭区间）的个数

方法一：归并排序

// 思路与转换：累加和 => 前缀和数组sum，
// i > j, 若sum[i] - sum[j] in [low, high], 则arr中[j + 1, j]区间和in[low, high]
// 利用归并merge过程两边有序，使比较不回退，降低时间复杂度
public int countOfRangeSum(int[] arr, int low, int high) {  
    if (arr == null) return 0;  
    int N = arr.length;  
    long[] sum = new long[N];  
    sum[0] = arr[0];  
    for (int i = 1; i < N; i++) {  
        sum[i] = sum[i - 1] + arr[i];  
    }  
    return process(sum, 0, N - 1, low, high);  
}  
  
private int process(long[] arr, int left, int right, int low, int high) {  
    if (left == right) {  
        return arr[left] >= low && arr[left] <= high ? 1 : 0;  
    }  
    int mid = left + ((right - left) >> 1);  
    int p1 = process(arr, left, mid, low, high);  
    int p2 = process(arr, mid + 1, right, low, high);  
    int p3 = merge(arr, left, mid, right, low, high);  
    return p1 + p2 + p3;  
}  
  
private int merge(long[] sum, int left, int mid, int right, int low, int high) {  
    int l1 = left, l2 = left, r = mid + 1;  
    long min, max;  
    // j < i; sum[i] - sum[j] in [low, high]  
    // sum[j] in [sum[i] - high, sum[i] - low]    
    // sum[j] 递增  [sum[i] - high, sum[i] - low] 递增， 不回退， 窗口  
    int ans = 0;  
    while (r <= right) {  
        min = sum[r] - high;  
        max = sum[r] - low;  
        while (l1 <= mid && sum[l1] < min) {  
            l1++;  
        }  
        while (l2 <= mid && sum[l2] <= max) {  
            l2++;  
        }  
        ans += l2 - l1;  
        r++;  
    }  
    // merge  
    int p1 = left;  
    int p2 = mid + 1;  
    long[] help = new long[right - left + 1];  
    int i = 0;  
    while (p1 <= mid && p2 <= right) {  
        help[i++] = sum[p1] <= sum[p2] ? sum[p1++] : sum[p2++];  
    }  
    while (p1 <= mid) {  
        help[i++] = sum[p1++];  
    }  
    while (p2 <= right) {  
        help[i++] = sum[p2++];  
    }  
    while (--i >= 0) {  
        sum[left + i] = help[i];  
    }  
    return ans;  
}

方法二：有序表

原题arr转换为前缀和数组sum，定义BST树节点定义如下，遍历sums, 设当前值为sum从BST中查找有多少在[sum-high, sum-low]

public class Node<K extends Comparable<K>> {
    K k;
    // 做子树的节点个数 + 右子树节点个数 + 自身（平衡用）
    int size;
    // 左子树K出现的次数 + 右子树K出现的次数 + 子树K出现的次数
    int all;

    Node<K> l, r;
}