Leetcode 506. Relative Ranks

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先上题目：

两个思路：

1. 暴力法。对输入的数组nums[]中的每一个元素都计算它在数组中的大小顺序rank，然后根据rank来计算输出，用switch case语句直接搞定。nums[]有n个元素，每个元素都需要与数组中的其他元素比较一次，时间复杂度是O(n^2) 测试通过时间116ms 代码见下：

int getDigit(int n);void itoa(int val, char* s, int digit);/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */char** findRelativeRanks(int* nums, int numsSize, int* returnSize) {    char** res;    int i, j;    int rank, digit;        *returnSize = numsSize;    res = (char**)malloc(sizeof(char*)*numsSize);        for(i = 0; i < numsSize; i++) {        rank = 1;        for(j = 0; j < numsSize; j++) {            if(i == j) {                continue;            }            else {                if(nums[i] < nums[j]) {                    rank++;                }            }        }                switch(rank) {            case 1:                res[i] = (char*)malloc(sizeof(char)*(strlen("Gold Medal")+1));                strcpy(res[i], "Gold Medal");                break;            case 2:                res[i] = (char*)malloc(sizeof(char)*(strlen("Silver Medal")+1));                strcpy(res[i], "Silver Medal");                break;            case 3:                res[i] = (char*)malloc(sizeof(char)*(strlen("Bronze Medal")+1));                strcpy(res[i], "Bronze Medal");                break;            default:                digit = getDigit(rank);                res[i] = (char*)malloc(sizeof(char)*(digit+1));                                res[i][digit] = '\0';                itoa(rank, res[i], digit);                                break;        }    }        return res;}int getDigit(int n) {    int d = 0;    while(n) {        n/=10;        d++;    }        return d;}void itoa(int val, char* s, int digit) {    int i, r;    for(i = 0; i < digit; i++) {        r = val%10;        s[digit-1-i] = r+'0';        val/=10;    }}

2. 额外利用一个数组scores[]，记录下nums[]中出现的所有分数及其对应的位置，即scores[]的key是nums[i]，value是对应的下标i。这种方法需要先找出nums[]中的最大值以确定scores[]的大小，赋值好scores[]之后，逆序判断各个元素的rank 找出最大值和赋值scores[]数组耗时O(n)，测试通过时间6ms 代码见下：

int getDigit(int n);void itoa(int val, char* s, int digit);char* getString(int n);/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */char** findRelativeRanks(int* nums, int numsSize, int* returnSize) {    char** res;    int i, max;    int rank;    int* scores;        *returnSize = numsSize;    res = (char**)malloc(sizeof(char*)*numsSize);        // find max nums[]    max = nums[0];    for(i = 1; i < numsSize; i++) {        if(max < nums[i]) {            max = nums[i];        }    }        // construct scores[]    scores = (int*)malloc(sizeof(int)*(max+1));    memset(scores, -1, sizeof(int)*(max+1));    for(i = 0; i < numsSize; i++) {        scores[nums[i]] = i;    }        // calculate rank    rank = 1;    for(i = max; i >= 0; i--) {        if(scores[i] != -1) {            res[scores[i]] = getString(rank);            rank++;        }    }        free(scores);    return res;}char* getString(int n) {    char* s;    int d;    switch(n) {        case 1:            s = (char*)malloc(sizeof(char)*(strlen("Gold Medal")+1));            strcpy(s, "Gold Medal");            break;        case 2:            s = (char*)malloc(sizeof(char)*(strlen("Silver Medal")+1));            strcpy(s, "Silver Medal");            break;        case 3:            s = (char*)malloc(sizeof(char)*(strlen("Bronze Medal")+1));            strcpy(s, "Bronze Medal");            break;        default:            d = getDigit(n);            s = (char*)malloc(sizeof(char)*(d+1));            s[d] = '\0';            itoa(n, s, d);            break;    }        return s;}int getDigit(int n) {    int d = 0;    while(n) {        n/=10;        d++;    }        return d;}void itoa(int val, char* s, int digit) {    int i, r;    for(i = 0; i < digit; i++) {        r = val%10;        s[digit-1-i] = r+'0';        val/=10;    }}

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