containsDuplicate.c

// 4.存在重复 方法一暴力方法
bool containsDuplicate(int* nums, int numsSize) {
    if (numsSize <= 1) {
        return false;
    }
    for (int i = 0; i < numsSize; i ++) {
        for (int j = i+1; j < numsSize; j++) {
            if (nums[i] == nums[j]) {
                return true;
            }
        }
    }
    return false;
}
// 方法二 哈希表，数组的下表当作key，value是这个数字出现的次数
bool containsDuplicate1(int* nums, int numsSize) {
    if (numsSize<2)
        return false;
    
    int min=nums[0],max=nums[0];
    for (int i=1;i<numsSize;i++){
        if (min==nums[i]||max==nums[i])
            return true;
        
        min=(nums[i]<min)?nums[i]:min;
        max=(nums[i]>max)?nums[i]:max;
    }
    
    int index[999999] = {0};
    for (int i=0;i<numsSize;i++){
        if(index[nums[i]-min]!=0)
            return true;
        else
            index[nums[i]-min]++;
    }
    return false;
}

int cmp(void const *a,void const *b);
// 方法三先排序后判断
bool containsDuplicate2(int* nums, int numsSize) {
    if ((numsSize == 0) || (numsSize ==1)) {
        return false;
    }
    qsort(nums, numsSize, sizeof(nums[0]), cmp);
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == nums[i+1]) {
            return true;
        }
    }
    return false;
}

int cmp(void const *a,void const *b) {
    return *(int*)a - *(int*)b;
}