交换函数
public static void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
1. 冒泡排序
public static void bubbleSort(int[] arr) {
for(int i = 0; i < arr.length - 1; i++) {
for(int j = 0; j < arr.length - i - 1; j++) {
if(arr[j] > arr[j + 1]) {
swap(arr, j, j + 1);
}
}
}
}
2. 选择排序
public static void selectSort(int[] arr) {
for(int i = 0; i < arr.length - 1; i++) {
int minIndex = i;
for(int j = i + 1; j < arr.length; j++) {
if(arr[j] < arr[minIndex])
minIndex = j;
}
swap(arr, i, minIndex);
}
}
3. 快速排序
public static void quickSort(int[] arr, int start, int end) {
if(start < end) {
int pivot = arr[start];
int low = start;
int high = end + 1;
while (true) {
while (low < end && arr[++low] <= pivot);
while (high > start && arr[--high] > pivot);
if(low < high)
swap(arr, low, high);
else
break;
}
swap(arr, start, high);
quickSort(arr, start, high - 1);
quickSort(arr, high + 1, end);
}
}
4.插入排序
public static void insertSort(int[] arr) {
for(int i = 1; i < arr.length; i++) {
int temp = arr[i];
int j;
for(j = i - 1; j >= 0 && arr[j] > temp; j--) {
arr[j + 1] = arr[j];
}
arr[j + 1] = temp;
}
}
5. 归并排序
const int maxn = 100;
void merge(int arr[], int l1, int r1, int l2, int r2) {
int temp[maxn];
int i = l1, j = l2, index = 0;
while(i <= r1 && j <= r2) {
if(arr[i] <= arr[j])
temp[index++] = arr[i++];
else
temp[index++] = arr[j++];
}
while(i <= r1)
temp[index++] = arr[i++];
while(j <= r2)
temp[index++] = arr[j++];
for(int i = 0; i < index; i++) {
arr[l1 + i] = temp[i];
}
}
void mergeSort(int arr[], int left, int right) {
if(left < right) {
int mid = left + (right - left) / 2;
mergeSort(arr, left, mid);
mergeSort(arr, mid + 1, right);
merge(arr, left, mid, mid + 1, right);
}
}
6. 二分查找
public static int find(int[] arr, int num) {
int start = 0, end = arr.length - 1;
while(start <= end) {
int mid = (start + end) / 2;
if(arr[mid] == num)
return mid;
else if(arr[mid] > num)
end = mid - 1;
else
start = mid + 1;
}
return -1;
}
7. 求递增序列中的第一个大于等于x的元素的值
int lower_bound(int A[], int left, int right, int x) {
int mid;
while(left < right) {
mid = left + (right - left) / 2;
if(A[mid] >= x)
right = mid;
else
left = mid + 1;
}
return left;
}
8. 求递增序列中的第一个大于x的元素的值
int upper_bound(int A[], int left, int right, int x) {
int mid;
while(left < right) {
mid = left + (right - left) / 2;
if(A[mid] > x)
right = mid;
else
left = mid + 1;
}
return left;
}
网友评论