leetcode学习(初级算法-数组)

C语言实现初级数组算法。

1.删除排序数组中的重复项

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2gy9m/实现代码:

int removeDuplicates(int* nums, int numsSize){
    int *i = nums, *end = nums + numsSize;
    while (++nums < end)
        if (*i != *nums) 
            *(++i) = *nums;
    return i - nums + numsSize + 1;
}

2.买卖股票的最佳时机 II

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2zsx1/实现代码:

int maxProfit(int *prices, int pricesSize){
    if (pricesSize < 2)
        return 0;
    int *end = prices + pricesSize - 1, min = *prices;
    int a=*prices,b=*(++prices),c;
    pricesSize = 0;
    while(prices=c)
                pricesSize+=b-min;
        }
        a=b;
        b=c;
        prices++;
    }
    if (b > a)
        return pricesSize + b - min;
    return pricesSize;
}

3.旋转数组

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2skh7/实现代码:

void reverse(int *nums, int l, int r)
{
    int n, len = (l + r) / 2, i, j = --r;
    for (i = l; i < len; i++, j--){
        n = *(nums + i);
        *(nums + i) = *(nums + j);
        *(nums + j) = n;
    }
}

void rotate(int *nums, int numsSize, int k)
{
    k = k % numsSize;
    reverse(nums, 0, numsSize);
    reverse(nums, 0, k);
    reverse(nums, k, numsSize);
}

4.存在重复元素

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x248f5/实现代码:

int *cmp(const void *a,const void *b){return *((int*)a)>*((int*)b)?0:1;}
bool containsDuplicate(int* nums, int numsSize){
    qsort(nums,numsSize,sizeof(int),cmp);
    int i;
    for(i=1;i

5.只出现一次的数字

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x21ib6/实现代码:

int singleNumber(int* nums, int numsSize){
    int *end=nums+numsSize,n=0;
    while(nums!=end){
        n^=*nums++;
    }
    return n;
}

6.两个数组的交集 II

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2y0c2/实现代码:

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int cmp(const void *a, const void *b) { return *((int *)a) < *((int *)b) ? 0 : 1; }
int *intersect(int *nums1, int nums1Size, int *nums2, int nums2Size, int *returnSize)
{
    qsort(nums1, nums1Size, sizeof(int), cmp);
    qsort(nums2, nums2Size, sizeof(int), cmp);
    int *retns = (int *)malloc((nums1Size < nums2Size ? nums1Size : nums2Size) * (sizeof(int)));
    int i = 0, j = 0, r = 0;
    for (; i < nums1Size; ++i){
        for (; j < nums2Size; ++j){
            if (nums1[i] == nums2[j]){
                retns[r++] = nums2[j++];
                break;
            }
            if (nums1[i] < nums2[j]){
                break;
            }
        }
    }
    *returnSize = r;
    return retns;
}

7.加一

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2cv1c/实现代码:

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int *plusOne(int *digits, int digitsSize, int *returnSize)
{
    //如果能在原数组的基础上修改,就不申请内存
    int i = digitsSize - 1;
    *returnSize = digitsSize;
    digits[i] += 1;
    for (; i > 0; --i){
        if (digits[i] > 9){
            digits[i] = 0;
            digits[i - 1] += 1;
        }else{
            break;
        }
    }
    int *ret;
    if (digits[0] > 9){
        digits[0] = 0;
        (*returnSize)++;
        ret = (int *)malloc((*returnSize) * sizeof(int));
        memcpy(ret + 1, digits, digitsSize * sizeof(int));
        ret[0] = 1;
        return ret;
    }
    return digits;
}

8.移动零

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2ba4i/实现代码:

void moveZeroes(int *nums, int numsSize)
{
    int i = 0, k = 0, t;
    for (; i < numsSize; ++i){
        if (0 != nums[i]){
            t = nums[i];
            nums[i] = 0;
            nums[k++] = t;
        }
    }
}

9.两数之和

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2jrse/实现代码:

int comp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int *twoSum(int *nums, int numsSize, int target, int *returnSize)
{
    int len = numsSize - 1, j = len, i = 0, n;
    int *ret = (int *)malloc(sizeof(int) * (2 + numsSize)), *num1;
    if (ret == NULL){
        *returnSize = 0;
        return NULL;
    }
    num1 = ret + 2;
    *returnSize = 2;
    memcpy(num1, nums, numsSize * sizeof(int));
    qsort(num1, numsSize, sizeof(int), comp);
    while (i < len && j > 0){
        n = target - num1[i];
        if (n < num1[j]) j--;
        else if (n > num1[j]) i++;
        else{
            ret[0] = num1[i];
            ret[1] = num1[j];
            break;
        }
    }
    for (i = 0, j = 0; i < numsSize && j < 2; ++i){
        if (nums[i] == ret[0]){
            ret[0] = i;
            j++;
        }
        else if (nums[i] == ret[1]){
            ret[1] = i;
            j++;
        }
    }
    return ret;
}

10.有效的数独

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2f9gg/实现代码:

bool isValidSudoku(char** board, int boardSize, int* boardColSize){
    int x, y[9] = {0}, b, bit, bi, i, j;
    for (i = 0; i < boardSize; ++i)
    {
        x = 0;
        if (i % 3 == 0) b = 0;
        for (j = 0; j < *boardColSize; ++j)
        {
            if (board[i][j] == '.') continue;
            bit = 1 << (board[i][j] - '0');
            bi = bit << (j / 3) * 9;
            if ((bit & x) || (bit & y[j]) || (bi & b))
                return false;
            x |= bit;
            y[j] |= bit;
            b |= bi;
        }
    }
    return true;
}

11.旋转图像

初级算法 - LeetBook - 力扣(LeetCode)全球极客挚爱的技术成长平台 (leetcode-cn.com)icon-default.png?t=L892https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/xnhhkv/实现代码:

void rotate(int **matrix, int matrixSize, int *matrixColSize)
{
    int i, j, e_j, e_i, tmp, ends = matrixSize - 1, len = ends;
    for (i = 0; i < matrixSize / 2; ++i, --len){
        for (j = ends - len; j < len; ++j){
            tmp = matrix[i][j];
            e_i = ends - i;
            e_j = ends - j;
            matrix[i][j] = matrix[e_j][i];
            matrix[e_j][i] = matrix[e_i][e_j];
            matrix[e_i][e_j] = matrix[j][e_i];
            matrix[j][e_i] = tmp;
        }
    }
}

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