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10:03:08",{"id":1084,"title":1085,"abbrlink":1086,"body":1087,"category":40,"cover":156,"date":156,"description":1093,"extension":1742,"mathjax":1743,"meta":1744,"navigation":1743,"path":691,"published":693,"readingMinutes":899,"seo":1747,"stem":1748,"sticky":156,"swiper_index":156,"tags":1749,"updated":156,"words":692,"__hash__":1750},"posts\u002Fposts\u002F2025\u002F2025-12-27-算法刷题-关于链表操作.md","2025-12-27-关于链表操作","40452",{"type":1088,"value":1089,"toc":1724},"minimark",[1090,1094,1099,1102,1121,1324,1327,1329,1332,1342,1401,1405,1409,1412,1420,1425,1431,1437,1449,1457,1465,1469,1474,1478,1486,1492,1499,1503,1508,1514,1520,1524,1529,1535,1541,1545,1550,1556,1562,1564,1566,1568,1572,1575,1578,1581,1584,1587,1590,1595,1598,1626,1629,1635,1641,1645,1648,1651,1661,1664,1691,1694,1709,1714,1720,1722],[1091,1092,1093],"p",{},"后悔数据结构当初没有好好学的第n天\n现在恶补，知识学爆",[1095,1096],"img",{"src":1097,"style":1098},"https:\u002F\u002Fpicx.zhimg.com\u002Fv2-1e47d2da440013ae7d4262f54c26dd52_720w.webp?source=d16d100b","height=100px",[1091,1100,1101],{},"基础操作：",[1103,1104,1105,1109,1112,1115,1118],"ul",{},[1106,1107,1108],"li",{},"查找元素：根据值查找节点位置",[1106,1110,1111],{},"指定位置插入：在特定位置插入新节点",[1106,1113,1114],{},"指定位置获取：获取特定位置的节点值",[1106,1116,1117],{},"指定位置删除：删除特定位置的节点",[1106,1119,1120],{},"获取长度：统计链表中节点的数量",[1122,1123,1124,1139],"table",{},[1125,1126,1127],"thead",{},[1128,1129,1130,1133,1136],"tr",{},[1131,1132],"th",{},[1131,1134,1135],{},"题目描述",[1131,1137,1138],{},"主要实现思路",[1140,1141,1142,1154,1170,1181,1192,1203,1214,1225,1236,1247,1258,1269,1280,1291,1302,1313],"tbody",{},[1128,1143,1144,1148,1151],{},[1145,1146,1147],"td",{},"BM1",[1145,1149,1150],{},"反转链表",[1145,1152,1153],{},"使用三指针（prev、curr、next）迭代反转：保存下一个节点，反转当前指向，移动指针。返回prev作为新头。",[1128,1155,1156,1159,1162],{},[1145,1157,1158],{},"BM2",[1145,1160,1161],{},"链表内指定区间反转",[1145,1163,1164,1165,1169],{},"引入虚拟头结点定位第m-1个节点（pre）。然后在",[1166,1167,1168],"span",{},"m,n","区间执行(n-m)次头插法（逐个将下一个节点插入pre后）。",[1128,1171,1172,1175,1178],{},[1145,1173,1174],{},"BM3",[1145,1176,1177],{},"链表中的节点每k个一组翻转",[1145,1179,1180],{},"每k个节点为一组，使用反转链表方法局部反转。若剩余不足k个，则保持原序。递归或迭代均可，推荐迭代分段处理。",[1128,1182,1183,1186,1189],{},[1145,1184,1185],{},"BM4",[1145,1187,1188],{},"合并两个排序的链表",[1145,1190,1191],{},"双指针模拟归并排序：比较两个链表当前节点值，小者接入新链表，移动对应指针。处理剩余部分。",[1128,1193,1194,1197,1200],{},[1145,1195,1196],{},"BM5",[1145,1198,1199],{},"合并k个已排序的链表",[1145,1201,1202],{},"使用小根堆（优先队列）维护k个链表头结点，每次弹出最小值并接入结果链表，同时推入其下一个节点。",[1128,1204,1205,1208,1211],{},[1145,1206,1207],{},"BM6",[1145,1209,1210],{},"判断链表中是否有环",[1145,1212,1213],{},"快慢指针（Floyd判圈算法）：fast每次走2步，slow走1步。若相遇则有环，否则无环。",[1128,1215,1216,1219,1222],{},[1145,1217,1218],{},"BM7",[1145,1220,1221],{},"链表中环的入口结点",[1145,1223,1224],{},"先用快慢指针相遇于环内某点，再令一指针从头启动，与慢指针同步移动，相遇处即环入口。",[1128,1226,1227,1230,1233],{},[1145,1228,1229],{},"BM8",[1145,1231,1232],{},"链表中倒数最后k个结点",[1145,1234,1235],{},"快慢指针：fast先走k步，然后slow与fast同步移动，至fast到尾时slow即为倒数第k个节点。",[1128,1237,1238,1241,1244],{},[1145,1239,1240],{},"BM9",[1145,1242,1243],{},"删除链表的倒数第n个节点",[1145,1245,1246],{},"同BM8定位倒数第n+1个节点（pre），然后pre.next = pre.next.next删除目标节点。注意头节点特殊处理。",[1128,1248,1249,1252,1255],{},[1145,1250,1251],{},"BM10",[1145,1253,1254],{},"两个链表的第一个公共结点",[1145,1256,1257],{},"双指针同步走：先计算长度差，长者先走差值步；或让指针走完一链表后换另一链表，总路程相等时相遇即公共节点。",[1128,1259,1260,1263,1266],{},[1145,1261,1262],{},"BM11",[1145,1264,1265],{},"链表相加(二)",[1145,1267,1268],{},"模拟加法从低位到高位（需先反转链表或用栈），处理进位。结果可能需反转回原序。",[1128,1270,1271,1274,1277],{},[1145,1272,1273],{},"BM12",[1145,1275,1276],{},"单链表的排序",[1145,1278,1279],{},"归并排序（自底向上）：分段合并有序子链表，或快慢指针找中点递归归并。时间O(n log n)。",[1128,1281,1282,1285,1288],{},[1145,1283,1284],{},"BM13",[1145,1286,1287],{},"判断一个链表是否为回文结构",[1145,1289,1290],{},"快慢指针找中点，反转后半部分，与前半部分逐节点比较值是否相等。恢复链表可选。",[1128,1292,1293,1296,1299],{},[1145,1294,1295],{},"BM14",[1145,1297,1298],{},"链表的奇偶重排",[1145,1300,1301],{},"分离奇偶位节点成两个链表（odd、even），然后even接odd尾部。注意偶数长度处理。",[1128,1303,1304,1307,1310],{},[1145,1305,1306],{},"BM15",[1145,1308,1309],{},"删除有序链表中重复的元素-I",[1145,1311,1312],{},"单指针遍历：若当前节点与下一节点值相同，跳过下一节点（保留首次出现）。",[1128,1314,1315,1318,1321],{},[1145,1316,1317],{},"BM16",[1145,1319,1320],{},"删除有序链表中重复的元素-II",[1145,1322,1323],{},"引入虚拟头结点，双指针或单指针遍历：若连续重复，跳过整个重复段（一个不留）。",[1325,1326],"hr",{},[1325,1328],{},[1091,1330,1331],{},"一般会给出一个最基础的链表",[1333,1334,1339],"pre",{"className":1335,"code":1337,"language":161,"meta":1338},[1336],"language-JavaScript","function ListNode(val, next) {\n    this.val = (val === undefined ? 0 : val);\n    this.next = (next === undefined ? null : next);\n}\n","",[1340,1341,1337],"code",{"__ignoreMap":1338},[1103,1343,1344,1359,1365,1371,1384],{},[1106,1345,1346,1350,1351],{},[1347,1348,1349],"strong",{},"节点包含两个成员","：\n",[1103,1352,1353,1356],{},[1106,1354,1355],{},"val：存储节点的值，通常为整数（int），题目中 |val| ≤ 1000。",[1106,1357,1358],{},"next：指向下一个节点的指针（引用），类型为同类 ListNode*（或 ListNode），初始可能为 NULL\u002Fnull\u002FNone",[1106,1360,1361,1364],{},[1347,1362,1363],{},"无哑头节点（dummy head）","：输入的 head 就是真实头结点（有有效值），除非题目特别说明",[1106,1366,1367,1370],{},[1347,1368,1369],{},"单向链表","：只能从头到尾遍历，无前向指针",[1106,1372,1373,1350,1376],{},[1347,1374,1375],{},"输入形式",[1103,1377,1378,1381],{},[1106,1379,1380],{},"函数签名通常为 ListNode* head（或类似），可能额外传入其他参数（如 m、n、k 等）",[1106,1382,1383],{},"空链表：head = NULL \u002F null \u002F None",[1106,1385,1386,1350,1389],{},[1347,1387,1388],{},"输出形式",[1103,1390,1391,1394],{},[1106,1392,1393],{},"大多数题目要求返回新的头结点（ListNode*）",[1106,1395,1396,1397,1400],{},"操作通常要求",[1347,1398,1399],{},"原地修改","，以满足空间复杂度 O(1)",[1402,1403,1404],"h2",{"id":1404},"常见操作",[1406,1407,1408],"h3",{"id":1408},"单链表常见操作的实现方法",[1091,1410,1411],{},"以下针对单链表（节点结构为 val 和 next）的几种常见操作，提供标准、高效的实现思路。所有操作均基于从头结点开始遍历，时间复杂度与空间复杂度分析清晰。假设节点定义如下（以 JavaScript 为例，其他语言类似）：",[1333,1413,1418],{"className":1414,"code":1416,"language":1417,"meta":1338},[1415],"language-ts","function ListNode(val, next) {\n  this.val = val\n  this.next = next || null\n}\n","ts",[1340,1419,1416],{"__ignoreMap":1338},[1421,1422,1424],"h4",{"id":1423},"_1-查找元素根据值查找节点位置返回位置或节点","1. 查找元素：根据值查找节点位置（返回位置或节点）",[1091,1426,1427,1430],{},[1347,1428,1429],{},"思路","：从头遍历，逐个比较节点值，直到找到匹配值或到达链表末尾。",[1091,1432,1433,1436],{},[1347,1434,1435],{},"实现要点","：",[1103,1438,1439,1446],{},[1106,1440,1441,1442,1445],{},"返回第一个匹配节点的",[1347,1443,1444],{},"位置","（从 1 开始）或节点本身。",[1106,1447,1448],{},"未找到返回 -1 或 null。",[1333,1450,1455],{"className":1451,"code":1453,"language":1454,"meta":1338},[1452],"language-js","function findNode(head, target) {\n  let pos = 1\n  let curr = head\n  while (curr !== null) {\n    if (curr.val === target) {\n      return pos \u002F\u002F 或 return curr; 返回节点本身\n    }\n    curr = curr.next\n    pos++\n  }\n  return -1 \u002F\u002F 未找到\n}\n","js",[1340,1456,1453],{"__ignoreMap":1338},[1103,1458,1459,1462],{},[1106,1460,1461],{},"时间复杂度：O(n)",[1106,1463,1464],{},"空间复杂度：O(1)",[1421,1466,1468],{"id":1467},"_2-指定位置插入在第-i-个位置插入新节点i-从-1-开始","2. 指定位置插入：在第 i 个位置插入新节点（i 从 1 开始）",[1091,1470,1471,1473],{},[1347,1472,1429],{},"：遍历到第 i-1 个节点，将新节点插入其后。特殊处理插入到头部（i=1）。",[1091,1475,1476,1436],{},[1347,1477,1435],{},[1103,1479,1480,1483],{},[1106,1481,1482],{},"若 i=1，新节点成为新头。",[1106,1484,1485],{},"若 i > 长度，插入失败或插入尾部（视题目要求）。",[1333,1487,1490],{"className":1488,"code":1489,"language":1417,"meta":1338},[1415],"function insertAtPosition(head, i, val) {\n  let newNode = new ListNode(val)\n  if (i === 1) {\n    newNode.next = head\n    return newNode \u002F\u002F 新头结点\n  }\n\n  let curr = head\n  for (let pos = 1; pos \u003C i - 1 && curr !== null; pos++) {\n    curr = curr.next\n  }\n  if (curr === null) return head \u002F\u002F i 超出范围，不插入\n\n  newNode.next = curr.next\n  curr.next = newNode\n  return head\n}\n",[1340,1491,1489],{"__ignoreMap":1338},[1103,1493,1494,1497],{},[1106,1495,1496],{},"时间复杂度：O(i) → 最坏 O(n)",[1106,1498,1464],{},[1421,1500,1502],{"id":1501},"_3-指定位置获取获取第-i-个节点的値i-从-1-开始","3. 指定位置获取：获取第 i 个节点的値（i 从 1 开始）",[1091,1504,1505,1507],{},[1347,1506,1429],{},"：遍历 i-1 步，直接返回当前节点的值。",[1333,1509,1512],{"className":1510,"code":1511,"language":1417,"meta":1338},[1415],"function getAtPosition(head, i) {\n  let curr = head\n  for (let pos = 1; pos \u003C i && curr !== null; pos++) {\n    curr = curr.next\n  }\n  return curr ? curr.val : null \u002F\u002F 未找到返回 null\n}\n",[1340,1513,1511],{"__ignoreMap":1338},[1103,1515,1516,1518],{},[1106,1517,1496],{},[1106,1519,1464],{},[1421,1521,1523],{"id":1522},"_4-指定位置删除删除第-i-个节点i-从-1-开始","4. 指定位置删除：删除第 i 个节点（i 从 1 开始）",[1091,1525,1526,1528],{},[1347,1527,1429],{},"：遍历到第 i-1 个节点，修改其 next 指向跳过第 i 个节点。特殊处理删除头结点。",[1333,1530,1533],{"className":1531,"code":1532,"language":1417,"meta":1338},[1415],"function deleteAtPosition(head, i) {\n  if (head === null) return null\n  if (i === 1) return head.next \u002F\u002F 删除头结点\n\n  let curr = head\n  for (let pos = 1; pos \u003C i - 1 && curr !== null; pos++) {\n    curr = curr.next\n  }\n  if (curr === null || curr.next === null) return head \u002F\u002F i 超出范围\n\n  curr.next = curr.next.next \u002F\u002F 跳过第 i 个节点\n  return head\n}\n",[1340,1534,1532],{"__ignoreMap":1338},[1103,1536,1537,1539],{},[1106,1538,1496],{},[1106,1540,1464],{},[1421,1542,1544],{"id":1543},"_5-获取链表长度节点数量","5. 获取链表长度（节点数量）",[1091,1546,1547,1549],{},[1347,1548,1429],{},"：遍历链表，累计计数。",[1333,1551,1554],{"className":1552,"code":1553,"language":1417,"meta":1338},[1415],"function getLength(head) {\n  let len = 0\n  let curr = head\n  while (curr !== null) {\n    len++\n    curr = curr.next\n  }\n  return len\n}\n",[1340,1555,1553],{"__ignoreMap":1338},[1103,1557,1558,1560],{},[1106,1559,1461],{},[1106,1561,1464],{},[1325,1563],{},[1325,1565],{},[1402,1567,1150],{"id":1150},[1406,1569,1571],{"id":1570},"bm1-全量反转链表","BM1 | 全量反转链表",[1091,1573,1574],{},"给定一个单链表的头结点pHead(该头节点是有值的，比如在下图，它的val是1)，长度为n，反转该链表后，返回新链表的表头。",[1091,1576,1577],{},"数据范围： 0≤n≤10000≤n≤1000",[1091,1579,1580],{},"要求：空间复杂度 O(1)O(1) ，时间复杂度 O(n)O(n) 。",[1091,1582,1583],{},"如当输入链表{1,2,3}时，",[1091,1585,1586],{},"经反转后，原链表变为{3,2,1}，所以对应的输出为{3,2,1}。",[1091,1588,1589],{},"以上转换过程如下图所示：",[1091,1591,1592],{},[1095,1593],{"alt":1338,"src":1594},"https:\u002F\u002Fuploadfiles.nowcoder.com\u002Fimages\u002F20211014\u002F423483716_1634206291971\u002F4A47A0DB6E60853DEDFCFDF08A5CA249",[1091,1596,1597],{},"解法：",[1103,1599,1600,1603,1606,1623],{},[1106,1601,1602],{},"初始化 prev 为 null（新链表的尾部）。",[1106,1604,1605],{},"current 从链表头节点开始。",[1106,1607,1608,1609],{},"在循环中：\n",[1103,1610,1611,1614,1617,1620],{},[1106,1612,1613],{},"暂存 next = current.next（避免指针丢失）。",[1106,1615,1616],{},"将 current.next 指向 prev（反转当前指针）。",[1106,1618,1619],{},"更新 prev = current（前移 prev）。",[1106,1621,1622],{},"更新 current = next（前移 current）。",[1106,1624,1625],{},"循环结束后，prev 指向原链表的尾节点（新头节点），更新 list.head = prev。",[1091,1627,1628],{},"其实就是三指针原地反转",[1091,1630,1631],{},[1095,1632],{"alt":1633,"src":1634},"alt text","https:\u002F\u002Fpic4.zhimg.com\u002Fv2-6a742659e12b185569b64a1f773bd993_b.webp",[1333,1636,1639],{"className":1637,"code":1638,"language":1417,"meta":1338},[1415],"export function ReverseList(head: ListNode): ListNode {\n  let prev = null\n  let curr = head\n\n  while (curr !== null) {\n    \u002F\u002F 保存下一个节点，防止断链\n    let next = curr.next \u002F\u002F 反转当前节点的指向\n\n    curr.next = prev \u002F\u002F 指针向前移动\n\n    prev = curr\n    curr = next\n  } \u002F\u002F prev 指向反转后的新头结点\n  return prev\n}\n",[1340,1640,1638],{"__ignoreMap":1338},[1406,1642,1644],{"id":1643},"bm2-反转链表部分区间","BM2 | 反转链表部分区间",[1091,1646,1647],{},"给定一个单链表的头结点 head，长度为 n，反转该链表从位置 m 到 n 的部分，返回反转后的链表。",[1091,1649,1650],{},"数据范围： 0≤m≤n≤n≤1000 ，链表中任意节点的值满足 |val|≤1000",[1091,1652,1653,1654,1657,1658,1660],{},"例如：",[1655,1656],"br",{},"\n给出的链表为 1→2→3→4→5→NULL1→2→3→4→5→NULL, m=2,n=4m=2,n=4,",[1655,1659],{},"\n返回 1→4→3→2→5→NULL1→4→3→2→5→NULL.",[1091,1662,1663],{},"要实现原地反转指定区间，需要：",[1665,1666,1667,1670,1673,1680],"ol",{},[1106,1668,1669],{},"找到反转区间的前一个节点（pre），即第 m-1 个节点。",[1106,1671,1672],{},"找到反转区间的最后一个节点（记为 end），即第 n 个节点。",[1106,1674,1675,1676,1679],{},"将 ",[1166,1677,1678],{},"m, n"," 区间使用经典链表反转方法进行原地反转。",[1106,1681,1682,1683],{},"正确连接反转后的区间与前后部分：\n",[1103,1684,1685,1688],{},[1106,1686,1687],{},"pre.next 指向反转后区间的新的头节点（原第 n 个节点）。",[1106,1689,1690],{},"反转后区间的尾节点（原第 m 个节点）指向 end.next。",[1091,1692,1693],{},"关键操作：",[1103,1695,1696,1699,1706],{},[1106,1697,1698],{},"先遍历定位到 pre 和反转区间的起始节点 start（第 m 个节点）。",[1106,1700,1701,1702,1705],{},"然后在 ",[1166,1703,1704],{},"start, end"," 区间内使用三指针迭代反转。",[1106,1707,1708],{},"最后调整指针连接。",[1091,1710,1711],{},[1095,1712],{"alt":1633,"src":1713},"https:\u002F\u002Fpica.zhimg.com\u002Fv2-88b21fbd767be1a41379e156251ad554_1440w.gif",[1333,1715,1718],{"className":1716,"code":1717,"language":1417,"meta":1338},[1415],"export function reverseBetween(head: ListNode | null, m: number, n: number): ListNode | null {\n  if (head === null || m === n) return head\n\n  const dummy = new ListNode(0)\n  dummy.next = head\n  let pre = dummy\n\n  \u002F\u002F 移动到第 m-1 个节点\n  for (let i = 0; i \u003C m - 1; i++) {\n    pre = pre.next!\n  }\n\n  let start = pre.next! \u002F\u002F 第 m 个节点（反转区间的原头部）\n  let then = start.next \u002F\u002F 第 m+1 个节点（待头插的节点）\n\n  \u002F\u002F 执行 (n - m) 次头插\n  for (let i = 0; i \u003C n - m; i++) {\n    start.next = then.next \u002F\u002F 从原区间摘除 then\n    then.next = pre.next \u002F\u002F then 插入 pre 之后（成为新头部）\n    pre.next = then \u002F\u002F 更新 pre 的 next\n    then = start.next \u002F\u002F 更新 then 为下一个待移动节点\n  }\n\n  return dummy.next\n}\n",[1340,1719,1717],{"__ignoreMap":1338},[1325,1721],{},[1325,1723],{},{"title":1338,"searchDepth":1725,"depth":1725,"links":1726},4,[1727,1738],{"id":1404,"depth":1728,"text":1404,"children":1729},2,[1730],{"id":1408,"depth":1731,"text":1408,"children":1732},3,[1733,1734,1735,1736,1737],{"id":1423,"depth":1725,"text":1424},{"id":1467,"depth":1725,"text":1468},{"id":1501,"depth":1725,"text":1502},{"id":1522,"depth":1725,"text":1523},{"id":1543,"depth":1725,"text":1544},{"id":1150,"depth":1728,"text":1150,"children":1739},[1740,1741],{"id":1570,"depth":1731,"text":1571},{"id":1643,"depth":1731,"text":1644},"md",true,{"uuid":1745,"slots":1746},"d15e1d00-e2fb-11f0-acce-8b717b55b534",{},{"title":1085,"description":1093},"posts\u002F2025\u002F2025-12-27-算法刷题-关于链表操作",[199,40,178],"21w37DJiDqDCpGsDSlPcskCzcWNWxrJCJiPGMz5iXAk",1790443287124]