在數組中找到第 k 小的數
【要求】

如果 arr 長度為 N，要求時間復雜度為 O(N)，額外空間復雜度為 O(1)。

public static int[] getMinKNumByHeap(int[] arr, int k){if(k < 1 || k > arr.length){return arr;}int[] kHeap = new int[k];for(int i = 0; i != k; i++){heapInsert(kHeap, arr[i], i);}for(int i = k; i != arr.length; i++){if(arr[i] < kHeap[0]){kHeap[0] = arr[i];heapify(kHeap, 0, k);}}return kHeap; }public static void heapInsert(int[] arr, int value, int index){arr[index] = value;while(index != 0){int parent = (index - 1) / 2;if(arr[parent] < arr[index]){swap(arr, parent, index);index = parent;}else{break;}} }public static void heapify(int[] arr, int index, int heapSize){int left = index * 2 + 1;int right = index * 2 + 2;int largest = index;while(left < heapSize){if(arr[left] > arr[index]){largest = left;}if(right < heapSize && arr[right] > arr[largest]){largest = right;}if(largest != index){swap(arr, largest, index);}else{break;}index = largest;left = index * 2 + 1;right = index * 2 + 2;} }public static int[] getMinKNumByBFPRT(int[] arr, int k){if(k < 1 || k > arr.length){return arr;}int minKth = getMinKthByBFPRT(arr, k);int[] res = new int[k];int index = 0;for(int i = 0; i != arr.length; i++){if(arr[i ] < minKth){res[index++] = arr[i];}}for(; index != res.length; index++){res[index] = minKth;}return res; } public static int getMinKthByBFPRT(int[] arr, int K){int[] copyArr = copyArray(arr);return select(copyArr, 0, copyArr.length - 1; k - 1); }public static int[] copyArray(int[] arr){int[] res = new int[arr.length];for(int i = 0; i != res.length; i++){res[i] = arr[i];}return res; }public static int select(int[] arr, int begin, int end, int i){if(begin == end){return arr[begin];}int pivot = medianoOfMedians(arr, begin, end);int[] pivotRange = partition(arr, begin, end, pivot);if(i >= pivotRange[0] && i <= pivotRange[1]){return arr[i];}else if(i < pivotRange[0]){return select(arr, begin, pivotRange[0] - 1, i);}else{return select(arr, pivotRange[1] + 1, end, i);} }public static int medianoOfMedians(int[] arr, int begin, int end){int num = end = begin + 1;int offset = num % 5 == 0 ? 0 : 1;int[] mArr = new int[num / 5 + offset];for(int i = 0; i < mArr.length; i++){int beginI = begin + i *5;int endI = begin + 4;mArr[i] = getMedian(arr, beginI, Math.min(end, endI));}return select(mArr, 0, mArr.length - 1, mArr.length / 2); }public static int[] partition(int[] arr, int begin, int end, int pivotValue){int small = begin - 1;int cur = begin;int big = end + 1;while(cur != big){if(arr[cur] < pivotValue){swap(arr, ++small, cur++);}else if(arr[cur] > pivotValue){swap(arr, cur, --big);}else{cur++;}}int[] range = new int[2];range[0] = small + 1;range[1] = big - 1;return range; }public static int getMedian(int[] arr, int begin, int end){insertionSort(arr, begin, end);int sum = end + begin;int mid = (sum / 2) + (sum % 2);return arr[mid]; }public static void insertionSort(int[] arr, int begin, int end){for(int i = begin + 1; i ！= end + 1; i++){for(int j = i; j != begin; j--){if(arr[j - 1] > arr[j]){swap(arr, j - 1, j);}else{break;}}} }public static void swap(int[] arr, index1, int index2){int tmp = arr[index];arr[index1] = arr[index2];arr[index2] = tmp; }

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