小和问题和逆序对问题
Posted dockerchen
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思路:使用归并排序,每一轮归并后都局部有序,可以利用这个,减少时间复杂度
小和问题
关键代码:
public static int mergeSort(int[] arr, int left, int right)
if (left == right)
return 0;
int mid = ((right - left) >> 1) + left;
return mergeSort(arr, left, mid) + mergeSort(arr, mid + 1, right) + merge(arr, left, mid, right);
public static int merge(int[] arr, int left, int mid, int right)
int[] help = new int[right - left + 1];
int i = 0;
int p1 = left;
int p2 = mid + 1;
int res = 0;
//每经过一轮归并,数据从小到大排序。在每一轮归并中,计算左边比右边小的数的总和
while (p1 <= mid && p2 <= right)
if (arr[p1] < arr[p2])
res += arr[p1] * (right - p2 + 1);
help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
while (p1 <= mid)
help[i++] = arr[p1++];
while (p2 <= right)
help[i++] = arr[p2++];
for (int j = 0; j < help.length; j++)
arr[left + j] = help[j];
return res;
逆序对问题
关键代码:
public static int mergeSort(int[] arr, int left, int right)
if (left == right)
return 0;
int mid = ((right - left) >> 1) + left;
return mergeSort(arr, left, mid) +mergeSort(arr, mid + 1, right) +merge(arr, left, mid, right);
public static int merge(int[] arr, int left, int mid, int right)
int[] help = new int[right - left + 1];
int i = 0;
int p1 = left;
int p2 = mid + 1;
int res=0;
//每经过一轮归并,数据从大到小排序
while (p1 <= mid && p2 <= right)
if (arr[p1] > arr[p2])
for (int j = p2; j <= right; j++)
res++;
System.out.println(arr[p1] + "," + arr[j]);
help[i++] = arr[p1] > arr[p2] ? arr[p1++] : arr[p2++];
while (p1 <= mid)
help[i++] = arr[p1++];
while (p2 <= right)
help[i++] = arr[p2++];
for (int j = 0; j < help.length; j++)
arr[left + j] = help[j];
return res;
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