Sort Integers by the Number of 1 Bits

快排，注意比较的方法，因为 bitCount 相同时也要排序。

class Solution {public int[] sortByBits(int[] arr) {if (arr == null || arr.length == 0) {return new int[0];}quickSort(arr, 0, arr.length-1);return arr;}private void quickSort(int[] arr, int l, int r) {if (l < r) {int mid = partition(arr, l, r);quickSort(arr, l, mid-1);quickSort(arr, mid+1, r);}}private int partition(int[] arr, int l, int r) {int key = arr[r];int i = l - 1;for (int j=l; j<r; j++) {if (smaller(arr[j], key)) {i++;int tmp = arr[i];arr[i] = arr[j];arr[j] = tmp;}}i++;int tmp = arr[i];arr[i] = arr[r];arr[r] = tmp;return i;}private boolean smaller(int i, int j) {int iCount = Integer.bitCount(i);int jCount = Integer.bitCount(j);if (iCount < jCount) {return true;}if (iCount == jCount) {return i < j;}return false;}
}

使用我自己写的 bitCount方法，是 10ms，使用 Integer.bitCount方法是 3ms。

private int bitCount(int i) {int result = 0;while (i != 0) {if ((i & 1) == 1) {result++;}i >>>= 1;}return result;
}

Integer.bitCount

public static int bitCount(int i) {i = i - ((i >>> 1) & 0x55555555);i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);i = (i + (i >>> 4)) & 0x0f0f0f0f;i = i + (i >>> 8);i = i + (i >>> 16);return i & 0x3f;
}

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