标签：merge turn -- 不能 fill res add from prope
You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
Example:
Input: [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
class Solution { public List<Integer> countSmaller(int[] nums) { List<Integer> list = new ArrayList(); int[] n = new int[nums.length]; if(nums.length == 0) return list; for(int i = 0; i < nums.length; i++){ for(int j = i + 1; j < nums.length; j++){ if(nums[j] < nums[i]) n[i]++; } list.add(n[i]); } return list; } }
尼玛，TLE了，咋不早说不能O（n²）
class Solution { public List<Integer> countSmaller(int[] nums) { if(nums == null || nums.length == 0) { return new ArrayList(); } int[][] arr = new int[nums.length][2]; for(int i = 0; i < arr.length; ++i) { arr[i] = new int[]{nums[i], i}; } int[] count = new int[nums.length]; ms(arr, count); List<Integer> ret = new ArrayList(); for(int i = 0; i < count.length; ++i) { ret.add(count[i]); } return ret; } private void ms(int[][] arr, int[] count) { int[][] temp = new int[arr.length][2]; Arrays.fill(temp, new int[2]); ms(arr, temp, 0, arr.length - 1, count); } private void ms(int[][] arr, int[][] temp, int left, int right, int[] count) { if(left == right) { return; } int mid = left + (right - left) / 2; ms(arr, temp, left, mid, count); ms(arr, temp, mid + 1, right, count); merge(arr, temp, left, mid + 1, right, count); } private void merge(int[][] arr, int[][] temp, int left, int mid, int right, int[] count) { int leftEnd = mid, tempPtr = left, tempSection = right - left + 1; int inverse = 0; while(left < leftEnd && mid <= right) { // changed from: if(arr[left] <= arr[mid]) { if(arr[left][0] <= arr[mid][0]) { temp[tempPtr] = arr[left++]; // added: count[temp[tempPtr][1]] += inverse; count[temp[tempPtr][1]] += inverse; }else { temp[tempPtr] = arr[mid++]; // added: inverse++; inverse++; } tempPtr++; } while(left < leftEnd) { // added: count[arr[left][1]] += inverse; count[arr[left][1]] += inverse; temp[tempPtr++] = arr[left++]; } while(mid <= right) { temp[tempPtr++] = arr[mid++]; } for(int i = 0; i < tempSection; ++i, --right) { arr[right] = temp[right]; } } }
mergesort
class Solution { int[] temp; public List<Integer> countSmaller(int[] nums) { List<Integer> res = new ArrayList<>(); int n = nums.length; temp = new int[n]; int[] indexes = new int[n]; for (int i = 0; i < n; i++) { indexes[i] = i; res.add(0); } mergeSort(0, n - 1, indexes, nums, res); return res; } void mergeSort(int start, int end, int[] indexes, int[] nums, List<Integer> res) { if (start >= end) { return; } int mid = (start + end) / 2; mergeSort(start, mid, indexes, nums, res); mergeSort(mid + 1, end, indexes, nums, res); merge(start, end, indexes, nums, res); } void merge(int start, int end, int[] indexes, int[] nums, List<Integer> res) { if (start >= end) { return; } int mid = (start + end) / 2; int j = mid + 1; int i = start; while (i <= mid) { while (j <= end && nums[indexes[i]] > nums[indexes[j]]) { j++; } res.set(indexes[i], res.get(indexes[i]) + j - mid - 1); i++; } i = start; j = mid + 1; int k = start; while (i <= mid && j <= end) { int a = nums[indexes[i]]; int b = nums[indexes[j]]; if (a < b) { temp[k++] = indexes[i]; i++; } else { temp[k++] = indexes[j]; j++; } } while (i <= mid) { temp[k++] = indexes[i++]; } while (j <= end) { temp[k++] = indexes[j++]; } for (k = start; k <= end; k++) { indexes[k] = temp[k]; } } }
315. Count of Smaller Numbers After Self
标签：merge turn -- 不能 fill res add from prope
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11774858.html
