算法

全排列

- (void)viewDidLoad {
    [super viewDidLoad];
    NSMutableArray *arrM = [NSMutableArray arrayWithObjects:@3,@2,@3, nil];
    [self testFunc:arrM index:0 num:(int)arrM.count];

}

//数组全排列
- (void)testFunc:(NSMutableArray *)array index:(int)index num:(int)num{
    if(index == num - 1){
        NSLog(@"%@",array);
    }else{
        for (int i = index; i<num; i++) {
                if([self needSwap:array index:i len:num]){//如果数组中没有重复数字这一步判断可以省略

            //将第i个元素交换至当前index下标处
                    [array exchangeObjectAtIndex:index withObjectAtIndex:i];
            //以递归的方式对剩下元素进行全排列
                    [self testFunc:array index:index +1 num:num];
            //将第i个元素交换回原处
                    [array exchangeObjectAtIndex:index withObjectAtIndex:i];
                }
            }
    }
}

//去重，防止数组中有重复数字
- (BOOL)needSwap:(NSMutableArray *)array index:(int)index len:(int)len{
    for (int i = index +1; i<len; i++) {
        if(array[index] == array[I]){
            return  NO;
        }
    }
    return YES;
}

//给定正整数n,计算出n个元素的集合{1,2，....，n}能够划分为多少个不同的非空集合

int Func(int n,int m){
  if(n<=2 || m==n || m==1){
      return 1;
  }else{
      return Func(n-1,m-1) + m*Func(n-1,m)
 }
}

//调用
int n;
int result;
for(int i = 1,i<n,i++){
    result = result +Func(n,i);
}

重建二叉树

public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);
        return root;
    }
    //前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}
    private TreeNode reConstructBinaryTree(int [] pre,int startPre,int endPre,int [] in,int startIn,int endIn) {
         
        if(startPre>endPre||startIn>endIn)
            return null;
        TreeNode root=new TreeNode(pre[startPre]);
         
        for(int i=startIn;i<=endIn;i++)
            if(in[i]==pre[startPre]){
                root.left=reConstructBinaryTree(pre,startPre+1,startPre+i-startIn,in,startIn,i-1);
                root.right=reConstructBinaryTree(pre,i-startIn+startPre+1,endPre,in,i+1,endIn);
                      break;
            }
                 
        return root;
    }
}

快排

- (void)viewDidLoad {
    [super viewDidLoad];
    // Do any additional setup after loading the view.
    NSMutableArray *sortArray = [NSMutableArray arrayWithObjects:@2,@1,@3,@5,@2,@9, nil];
    [self quickSortTest:sortArray leftIndex:0 rightIndex:sortArray.count-1];
    NSLog(@"%@",sortArray);
}
- (void)quickSortTest:(NSMutableArray *)array leftIndex:(NSInteger)leftIndex rightIndex:(NSInteger)rightIndex{
    if(leftIndex>=rightIndex){
        return;
    }
    NSInteger i = leftIndex;
    NSInteger j = rightIndex;
    NSInteger key = [array[leftIndex] integerValue];
    while (i<j) {
        while (i<j && key<=[array[j] integerValue]) {
            j--;
        }
        if(i!=j){
            [array exchangeObjectAtIndex:i withObjectAtIndex:j];
        }
        while (i<j && key>=[array[i] integerValue]) {
            I++;
        }
        if(i!=j){
            [array exchangeObjectAtIndex:i withObjectAtIndex:j];
        }
    }
    [self quickSortTest:array leftIndex:leftIndex rightIndex:i-1];
    [self quickSortTest:array leftIndex:i+1 rightIndex:rightIndex];
}

链表反转

struct Node{
    int data;
    struct Node *next;
};
@interface ViewController ()

@end

@implementation ViewController
- (void)viewDidLoad {
    [super viewDidLoad];

}
struct Node *reverseList(struct Node *head){
    struct Node *p = head;
    struct Node *newH = NULL;
    while (p != NULL) {
        struct Node *temp = p->next;
        p->next = newH;
        newH = p;
        p = temp;
    }
    return newH;
}

有序链表的合并

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
//递归
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {

        if (l1 == NULL) {
            return l2;
        }
        if (l2 == NULL) {
            return l1;
        }
        ListNode *mergeNode = NULL;
        if (l1->val < l2->val) {
            mergeNode = l1;
            mergeNode->next = mergeTwoLists(l1->next, l2);
        } else {
            mergeNode = l2;
            mergeNode->next = mergeTwoLists(l1, l2->next);
        }
        return mergeNode;
    }
//循环

public class Solution {

if(l1 == null) {

return l2;

}

if(l2 == null) {

return l1;

}

ListNode dummy = new ListNode(0);

ListNode cur = dummy;

while(l1 != null&& l2 != null) {

if(l1.val < l2.val) {

cur.next = l1;

cur = cur.next;

l1 = l1.next;

} else{

cur.next = l2;

cur = cur.next;

l2 = l2.next;

}

}

if(l1 != null) {

cur.next = l1;

}

if(l2 != null) {

cur.next = l2;

}

return dummy.next;

}

二叉树的反转
步骤简述为：

public TreeNode invertTree(TreeNode root) {
        if(root==null){
            return null;
        }
        invertTree(root.left);
        invertTree(root.right);

        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        return root;
    }

字符串转数组
java

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