计数排序and基数排序

Posted 2020-10-07 wheleetcode

1 计数排序，稳定    复杂度o(k + n)

    public static int[] countingSort(int[] nums) {
        int n = nums.length;
        int k = 0;
        for (int i = 0; i < n; i++) {
            k = Math.max(k, nums[i]);
        }
        int[] count = new int[k + 1];
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            count[nums[i]]++;
        }
        for (int i = 1; i <= k; i++) {
            count[i] += count[i - 1];
        }
        for (int i = n - 1; i >= 0; i--) {
            res[--count[nums[i]]] = nums[i];
        }
        return res;
    }

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2 基数排序 需要稳定排序  有n个d位数，每一位有k个取值，复杂度为d（n + k）

    public static int[] radixSort(int[] nums) {
        int max = 0;
        int n = nums.length;
        for (int i = 0; i < nums.length; i++) {
            max = Math.max(max, nums[i]);
        }
        int exp = 1;
        int[] cur = new int[n];
        while (max / exp > 0) {
            int[] count = new int[10];            
            for (int i = 0; i < n; i++) {
                count[(nums[i] / exp) % 10]++;
            }
            for (int i = 1; i < 10; i++) {
                count[i] += count[i - 1];
            }
            for (int i = n - 1; i >= 0; i--) {
                cur[--count[(nums[i] / exp) % 10]] = nums[i];
            }
            for (int i = 0; i < n; i++) {
                nums[i] = cur[i];
            }
            exp *= 10;
        }
        return nums;
    }

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3 桶排序