LeetCode090 Subsets II

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详细见：leetcode.com/problems/subsets-ii

Java Solution: github

package leetcode;import java.util.ArrayList;import java.util.Arrays;import java.util.Iterator;import java.util.LinkedList;import java.util.List;/* * Given a collection of integers that might contain duplicates, nums,  * return all possible subsets.Note: The solution set must not contain duplicate subsets.For example,If nums = [1,2,2], a solution is:[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []] */public class P090_SubsetsII {public static void main(String[] args) {int[] nums = null;nums = new int[] {1, 1, 2, 2, 3, 3};List<List<Integer>> ans = new Solution().subsetsWithDup(nums);Iterator<List<Integer>> it = ans.iterator();while (it.hasNext()) {tools.Utils.B_打印List_Integer_OneLine(it.next());}}/* * AC * 无论何时，都需要耐心分析边界问题。 * 3 ms */static class Solution {List<List<Integer>> ans = new LinkedList<>();int[] first_same, last_same, record, nums;    public List<List<Integer>> subsetsWithDup(int[] nums) {    ans.add(new LinkedList<Integer>());    if (nums == null || nums.length == 0) {    return ans;    }    Arrays.sort(nums);    this.nums = nums;    first_same = new int[nums.length];    last_same = new int[nums.length];    record = new int[nums.length];    for (int i = 0; i != nums.length; i ++) {    if (0 == i || nums[i - 1] != nums[i]) {    first_same[i] = i;    } else {    first_same[i] = first_same[i - 1];    }    }    for (int i = nums.length - 1; i > - 1; i --) {    if (i == nums.length - 1 || nums[i] != nums[i + 1]) {    last_same[i] = i;    } else {    last_same[i] = last_same[i + 1];    }    }//    tools.Utils.printArray(first_same, 10);//    tools.Utils.printArray(last_same, 10);    for (int len = 1; len <= nums.length; len ++) {    search(len, 0, 0);    }        return ans;    }private void search(int len, int len_now, int sti) {if (len_now == len) {List<Integer> temp = new ArrayList<Integer>(len);for (int i = 0; i != len; i ++) {temp.add(record[i]);}ans.add(temp);return;}for (int i = sti; i < nums.length; i ++) {if (first_same[i] == last_same[i]) {record[len_now] = nums[i];search(len, len_now + 1, i + 1);} else {int count_same = last_same[i] - first_same[i];for (int same_now = 0; same_now <= count_same; same_now ++) {if (len_now + same_now >= len) {break;}//if (same_now != 0) {record[len_now + same_now] = nums[i];//}search(len, len_now + same_now + 1, last_same[i] + 1);}i += count_same;}}}}}

C Solution: github

/*    url: leetcode.com/problems/subsets-ii    AC 3ms 70.59%*/#include <stdio.h>#include <stdlib.h>typedef int* T;typedef struct al sal;typedef struct al * pal;struct al {    int capacity;    int size;    T* arr;};pal al_init(int capacity) {    pal l = (pal) malloc(sizeof(sal));    if (capacity < 1) return NULL;    l->arr = (T*) malloc(sizeof(T) * capacity);    l->capacity = capacity;    l->size = 0;    return l;}void al_expand_capacity(pal l) {    T* new_arr = (T*) malloc(sizeof(T) * (l->capacity * 2 + 1));    int i = 0;    for (i = 0; i < l->capacity; i ++)        new_arr[i] = l->arr[i];    free(l->arr);    l->arr = new_arr;    l->capacity = l->capacity * 2 + 1;}void al_add_last(pal l, T v) {    if (l->capacity == l->size) al_expand_capacity(l);    l->arr[l->size] = v;    l->size ++;}T* al_convert_to_array_free_l(pal l) {    T* arr = l->arr;    free(l);    return arr;}void al_free(pal l) {    int i = 0;    for (i = 0; i < l->size; i ++)        free(l->arr[i]);    free(l->arr);    free(l);}int partition(int* n, int i, int j) {    int s = n[i];    while (i < j) {        while (i < j && n[j] >= s) j --;        n[i] = n[j];        while (i < j && n[i] <= s) i ++;        n[j] = n[i];    }    n[i] = s;    return i;}void quick_sort(int* n, int i, int j) {    int p = 0;    if (i+1 < j) {        p = partition(n, i, j-1);        quick_sort(n, i, p);        quick_sort(n, p+1, j);    }}int* arr_copy(int* s, int sn) {    int* t = (int*) malloc(sizeof(int) * sn);    int i = 0;    for (i = 0; i < sn; i ++) {        t[i] = s[i];    }    return t;}int* arr_int(int val) {    int* t = (int*) malloc(sizeof(int));    t[0] = val;    return t;}void search(pal l, pal ln, int* n, int ni, int nn, int* s, int si, int sn, int sign) {    int i = 0;    if (si == sn && ni == nn) {        al_add_last(l, arr_copy(s, sn));        al_add_last(ln, arr_int(sn));        return;    }    if (ni >= nn || si > sn) return;    if (! (sign && ni != 0 && n[ni-1] == n[ni])) {        search(l, ln, n, ni+1, nn, s, si, sn, 0);    }    if (si < sn) s[si] = n[ni];    search(l, ln, n, ni+1, nn, s, si+1, sn, 1);}int** subsetsWithDup(int* n, int nn, int** cn, int* rn) {    pal l = NULL, ln = NULL;    int ni = 0, *s = NULL, si = 0;    if (n == NULL || nn < 1) return NULL;    quick_sort(n, 0, nn);    l = al_init(16);    ln = al_init(16);    s = (int*) malloc(sizeof(int) * nn);    for (ni = 0; ni <= nn; ni ++) {        search(l, ln, n, 0, nn, s, 0, ni, 0);    }    *rn = l->size;    *cn = (int*) malloc(sizeof(int) * *rn);    for (si = 0; si < *rn; si ++)        (*cn)[si] = ln->arr[si][0];   al_free(ln);   return al_convert_to_array_free_l(l);}int main() {    int n[] = {1,4,3,5,4,4,7,7,8,0};    int nn = 2;    int rn = 0;    int* cn = NULL;    int** a = subsetsWithDup(n, nn, &cn, &rn);    int i = 0, j = 0;    for (i = 0; i < rn; i ++) {        for (j = 0; j < cn[i]; j ++) {            printf("%d ", a[i][j]);        }        printf("\r\n");        free(a[i]);       }    free(a);}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/subsets-ii    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年4月22日    @details:    Solution: 65ms 71.66%'''class Solution(object):    def search(self, n, ni, nn, r, ri, rn, iu, a):        if ni == nn and ri == rn:            a.append(list(r))        if ni >= nn or ri > rn: return        if not (ni != 0 and iu and n[ni-1]==n[ni]):            self.search(n, ni+1, nn, r, ri, rn, False, a)        if ri < rn: r[ri] = n[ni]        self.search(n, ni+1, nn, r, ri+1, rn, True, a)            def subsetsWithDup(self, n):        """        :type n: List[int]        :rtype: List[List[int]]        """        nn = 0 if n == None else len(n)        if nn == 0: return []        n.sort(key=None, reverse=False)        a = []        for rn in range(nn+1):            r = [0]*rn            self.search(n, 0, nn, r, 0, rn, False, a)        return aif __name__ == "__main__":    a = Solution().subsetsWithDup([1,2,2,2])    print(a)

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