1、二分查找
已知一个有序数组, 和一个 key, 要求从数组中找到 key 对应的索引位置
int binaryFind(int *arr,int len,int key){ int min=0,max=len-1,mid; while (min <= max) { mid = (min+max)/2; if (key < arr[mid]) { max = mid-1; } else if (key > arr[mid]) { min = mid+1; } else { return mid; } } return -1;}2、字符串反转
- (void)strReverseTest{ char str[] = "32415"; int len = strlen(str); for (int i=0; i<(len+1)/2; i++) { char temp = str; str = str[len-1-i]; str[len-1-i] = temp; } NSLog(@"%s",str); // C中打印数组得for循环,不方便}3、有序数组归并
将有序数组 {1,4,6,7,9} 和 {2,3,5,6,8,9,10,11,12} 归并为 {1,2,3,4,5,6,6,7,8,9,9,10,11,12}
- (void)arrayMergeTest { int a[] = {1,4,6,7,9}; int b[] = {2,3,5,6,8,9,10,11,12}; int lenA=5,lenB=9,result[lenA+lenB],i=0,j=0,k=0; while (i<lenA && j<lenB) { if (a<b[j]) { result[k++] = a[i++]; } else { result[k++] = b[j++]; } } while (i<lenA) { result[k++] = a[i++]; } while (j<lenB) { result[k++] = b[j++]; } NSLog(@"end");}4、查找两个子视图的共同父视图
- (NSMutableArray*)findSuperview UIView*)view { NSMutableArray *array = [NSMutableArray array]; UIView*spView = view.superview; while (spView) { [array addObject:spView]; spView = spView.superview; } return array;}- (NSMutableArray*)findCommonSuperviewUIView*)view1 view2UIView*)view2{ NSMutableArray *array = [NSMutableArray array]; NSArray *superviews1 = [self findSuperview:view1]; NSArray *superviews2 = [self findSuperview:view2]; int len = MIN(superviews1.count,superviews2.count); for (int i=0; i<len; i++) { UIView *super1 = superviews1[superviews1.count-1-i]; UIView *super2 = superviews2[superviews2.count-1-i]; if (super1 == super2) { [array addObject:super1]; } else { break; } } return array;}5、求无序数组中的中位数
中位数：数组中最中间的谁人数或最中间的那两个数的匀称值
思绪：先排序，再求中位数
float midValue(int arr[], int len){ float mid = (len%2 == 0) ? (arr[len/2] + arr[len/2-1])/2.0 : arr[len/2]; return mid;}void midValueSort(int arr[], int len){ for (int i=0; i<len; i++) { for (int j=0; j<len-1-i; j++) { if (arr[j] > arr[j+1]) { int temp = arr[j]; arr[j] = arr[j+1]; arr[j+1]=temp; } } }}- (void)midValueTest { int arr[] = {5,1,3,6,100,8}; midValueSort(arr,6); float mid = midValue(arr, 6); NSLog(@"%.1f",mid); }6、Hash算法
例:在一个字符串中找到第一个只出现一次的字符[焦颔首脑:天生一个数组,数组中存储每个字符出现的次数]
char findFirstChar(char cha[], int len) { char result = '\0'; int arr[128]; for (int i = 0; i < 128; i++) { arr = 0; } for (int i = 0; i < len; i++) { arr[cha]++; } for (int i = 0; i < len; i++) { if (arr[cha] == 1) { result = cha; break; } } return result;}- (void)findFirstCharTest { char cha[] = "aabbcdddf"; char result = findFirstChar(cha, 9); NSLog(@"%c",result);} |
