[leetcode-60]Permutation Sequence(C)

问题描述：

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,

We get the following sequence (ie, for n = 3):

“123”

“132”

“213”

“231”

“312”

“321”

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

分析：看到这道题，我突然想到了之前做的leetcode中next permutation的那题，结果我直接复制进来，然后居然ac了，只是耗时太长了（192ms）。既然选择出满足条件的耗时太长，那么只能是构造出满足条件的数据了。

在构造数据时，我们可以看到，第j个数右边有（n-j)!种选择。即根据这样方式构造出数据来。代码如2：

代码如下：192ms

[code]void reverse(int *nums,int start,int end){
    int left = start;
    int right = end-1;

    int tmp;
    while(left<right){
        tmp = nums[left];
        nums[left] = nums[right];
        nums[right] = tmp;
        left++;
        right--;
    }
}
void nextPermutation(int* nums, int numsSize) {
    if(numsSize<2)
        return;
    int i,j;
    for(i = numsSize-2;i>=0;i--){
        if(nums[i]<nums[i+1])
            break;
    }
    if(i<0){
        reverse(nums,0,numsSize);
        return;
    }
    int min = nums[i+1];
    int index = i+1;
    for(j = i+2;j<numsSize;j++){
        if(nums[j]>nums[i]&&nums[j]<=min){
            min = nums[j];
            index = j;
        }
    }

    nums[index] = nums[i];
    nums[i] = min;

    reverse(nums,i+1,numsSize);
}
char* getPermutation(int n, int k) {
    int *nums = (int *)malloc(sizeof(int)*n);
    char *res = (char *)malloc(n+1);
    int i;
    for(i = 0;i<n;i++){
        nums[i] = i+1;
    }
    for(i = 1;i<k;i++){
        nextPermutation(nums,n);
    }
    res
 = '\0';
    for(i = 0;i<n;i++)
        res[i] = nums[i]+'0';

    return res;
}

写的不大优美，主要维护三个数组，第一个是nums，其中nums[j] = (n-j-1)*nums[j+1];第二个是origin，其中origin[j] = 1，表示，数j尚未被使用，0表示已经被使用。第三个是res,表示返回值。

代码如下：0ms

[code]void initNums(int *nums,int n){
    nums[n-1] = 1;

    int i;
    for(i = n-2;i>=0;i--)
        nums[i] = (n-i-1)*nums[i+1];

    nums[n-1] = 0;
}
void initOrigin(char *origin,int n){
    int i;
    for(i = 0;i<n;i++)
        origin[i] = 1;
}
char* getPermutation(int n, int k) {
    int *nums = (int *)malloc(sizeof(int)*n);
    char* res = (char *)malloc(sizeof(char)*(n+1));
    char* origin = (char *)malloc(sizeof(char)*(n+1));
    int resIndex = 0;
    int i;

    initNums(nums,n);
    initOrigin(origin,n+1);
    k--;

    for(i = n-1;i>=0;i--){
        if(nums[i]>k)
            break;
    }

    int index = i;
    for(i = 0;i<=index;i++){
        res[resIndex++] = i+1+'0';
        origin[i+1] = 0;//代表不可用
    }

    int offset = 1;
    while(k){
        int count = 0;
        while(k>=nums[index+offset]){
            count++;
            k=k-nums[index+offset];
        }
        offset++;
        for(i = index+2;i<=n;i++){
            if(origin[i]==0)
                continue;
            if(!count){
                res[resIndex++] = i+'0';
                origin[i] = 0;
                break;
            }
            count--;
        }
    }
    //deal with the res
    for(i = index+2;i<=n;i++){
        if(origin[i]!=0){
            res[resIndex++] = i+'0';
            origin[i] = 0;
        }
    }

    res[resIndex++] = '\0';
    return res;
}