O(n)
,n是链表的数量,最差的情况下每个节点都要遍历。空间复杂度O(n)
,n是存储遍历过的节点的map或者setjs:
var hasCycle = (head) => {
let map = new Map();
while (head) {
if (map.has(head)) return true;//如果当前节点在map中存在就说明有环
map.set(head, true);//否则就加入map
head = head.next;//迭代节点
}
return false;//循环完成发现没有重复节点,说明没环
};
java:
public class Solution {
public boolean hasCycle(ListNode head) {
Set<ListNode> seen = new HashSet<ListNode>();
while (head != null) {
if (!seen.add(head)) {
return true;
}
head = head.next;
}
return false;
}
}
O(n)
,空间复杂度O(1)
js:
var hasCycle = function (head) {
//设置快慢指针
let slow = head;
let fast = head;
//如果没有环,则快指针会抵达终点,否则继续移动双指针
while (fast && fast.next) {
slow = slow.next;
fast = fast.next.next;
//快慢指针相遇,说明含有环
if (slow == fast) {
return true;
}
}
return false;
};
java:
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if (fast == slow) {
return true;
}
}
return false;
}
}
O(n)
,空间复杂度O(n)
js:
var detectCycle = function(head) {
const visited = new Set();
while (head !== null) {//终止条件,如果没有环 跳出循环
if (visited.has(head)) {//如果存在重复的节点,这个节点就是入环节点
return head;
}
visited.add(head);//将节点加入set中
head = head.next;
}
return null;
};
java:
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode pos = head;
Set<ListNode> visited = new HashSet<ListNode>();
while (pos != null) {
if (visited.contains(pos)) {
return pos;
} else {
visited.add(pos);
}
pos = pos.next;
}
return null;
}
}
ds_175
O(n)
,空间复杂度O(1)
js:
var detectCycle = function(head) {
if (head === null) {
return null;
}
let slow = head, fast = head;
while (fast !== null) {
slow = slow.next;//慢指针移动两步,快指针移动一步
if (fast.next !== null) {
fast = fast.next.next;
} else {
return null;//如果没有环 之间返回null
}
if (fast === slow) {//有环
let fast = head;
//快指针指向头节点,然后每次快慢指针各走一步直到相遇,相遇的节点就是入环节点
while (fast !== slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
}
return null;
};
java:
public class Solution {
public ListNode detectCycle(ListNode head) {
if (head == null) {
return null;
}
ListNode slow = head, fast = head;
while (fast != null) {
slow = slow.next;
if (fast.next != null) {
fast = fast.next.next;
} else {
return null;
}
if (fast == slow) {
ListNode fast1 = head;
while (fast1 != slow) {
fast1 = fast1.next;
slow = slow.next;
}
return fast1;
}
}
return null;
}
}
方法1.暴力求解,对于三个数字,循环3次,分别计算和,时间复杂度O(n^3)
方法2.c=-(a+b): 确定了a和b,那就可以想两数之和一样,在map中寻找-(a+b)
,减少一层循环,时间复杂度O(n^2)
,空间复杂度O(n)
。
方法3.排序然后查找
L = i+1
,和R = nums.length-1
,如果和sum=nums[i] + nums[L] + nums[R]
小于0,则向右移动左指针,如果sum大于0,则左移右指针,如果sum等于0,则正好找到了这3个数,然后在尝试L++
,R--
,继续寻找中间是否有三个数之和等于0,注意在循环的过程中遇见相同的三个数需要去重。O(n^2)
,n为数组的长度。空间复杂度O(logn)
,即排序所需要的空间js:
var threeSum = function(nums) {
let ans = [];
const len = nums.length;
if(nums == null || len < 3) return ans;//数组的长度大于3
nums.sort((a, b) => a - b); // 排序
for (let i = 0; i < len ; i++) {
if(nums[i] > 0) break; // 如果当前数字大于0,则三数之和一定大于0,所以结束循环
if(i > 0 && nums[i] == nums[i-1]) continue; // 去重
let L = i+1;
let R = len-1;
while(L < R){//虽然里面还有两个循环,但是整体的L和R移动的时间内复杂度还是o(n)
const sum = nums[i] + nums[L] + nums[R];
if(sum == 0){
ans.push([nums[i],nums[L],nums[R]]);
while (L<R && nums[L] == nums[L+1]) L++; // 去重
while (L<R && nums[R] == nums[R-1]) R--; // 去重
L++;
R--;
}
else if (sum < 0) L++;
else if (sum > 0) R--;
}
}
return ans;
};
java:
class Solution {
public static List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> ans = new ArrayList();
int len = nums.length;
if(nums == null || len < 3) return ans;
Arrays.sort(nums);
for (int i = 0; i < len ; i++) {
if(nums[i] > 0) break; // 如果当前数字大于0,则三数之和一定大于0,所以结束循环
if(i > 0 && nums[i] == nums[i-1]) continue; // 去重
int L = i+1;
int R = len-1;
while(L < R){
int sum = nums[i] + nums[L] + nums[R];
if(sum == 0){
ans.add(Arrays.asList(nums[i],nums[L],nums[R]));
while (L<R && nums[L] == nums[L+1]) L++; // 去重
while (L<R && nums[R] == nums[R-1]) R--; // 去重
L++;
R--;
}
else if (sum < 0) L++;
else if (sum > 0) R--;
}
}
return ans;
}
}
O(n)
,n是数组height的长度,遍历一次。空间复杂度O(1)
js:
var maxArea = function(height) {
let max = 0;
for (let i = 0, j = height.length - 1; i < j;) {//双指针i,j循环height数组
//i,j较小的那个先向内移动 如果高的指针先移动,那肯定不如当前的面积大
const minHeight = height[i] < height[j] ? height[i++] : height[j--];
const area = (j - i + 1) * minHeight;//计算面积
max = Math.max(max, area);//更新最大面积
}
return max;
};
java:
class Solution {
public int maxArea(int[] height) {
int i = 0, j = height.length - 1, max = 0;
while(i < j) {
max = height[i] < height[j] ?
Math.max(max, (j - i) * height[i++]):
Math.max(max, (j - i) * height[j--]);
}
return max;
}
}
O(m+n)
,m、n分别是两个链表的长度。空间复杂度O(m)
js:
var getIntersectionNode = function(headA, headB) {
const visited = new Set();
let temp = headA;
while (temp !== null) {//将链表A存入set中
visited.add(temp);
temp = temp.next;
}
temp = headB;
while (temp !== null) {
if (visited.has(temp)) {//第一个相同的节点就是重合的节点
return temp;
}
temp = temp.next;
}
return null;
};
Java:
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
Set<ListNode> visited = new HashSet<ListNode>();
ListNode temp = headA;
while (temp != null) {
visited.add(temp);
temp = temp.next;
}
temp = headB;
while (temp != null) {
if (visited.contains(temp)) {
return temp;
}
temp = temp.next;
}
return null;
}
}
pA == pB
时就是交点,因为两个指针移动的步数一样O(m+n)
,m、n分别是两个链表的长度。空间复杂度O(1)
js:
var getIntersectionNode = function(headA, headB) {
if (headA === null || headB === null) {
return null;
}
let pA = headA, pB = headB;
while (pA !== pB) {
pA = pA === null ? headB : pA.next;//链表A循环结束就循环链表B
pB = pB === null ? headA : pB.next;//链表A循环结束就循环链表B
}
return pA;//当pA == pB时就是交点
};
java:
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) {
return null;
}
ListNode pA = headA, pB = headB;
while (pA != pB) {
pA = pA == null ? headB : pA.next;
pB = pB == null ? headA : pB.next;
}
return pA;
}
}
ds_185
O(n)
,空间复杂度O(1)
js:
var middleNode = function(head) {
slow = fast = head;
while (fast && fast.next) {//快慢指针遍历,直到快指针到达最后
slow = slow.next;
fast = fast.next.next;
}
return slow;
};
java:
class Solution {
public ListNode middleNode(ListNode head) {
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。