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本文隶属于专栏《LeetCode 刷题汇总》，该专栏为笔者原创，引用请注明来源，不足和错误之处请在评论区帮忙指出，谢谢！

本专栏目录结构请见LeetCode 刷题汇总

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algorithm

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131. 分割回文串

Java: Backtracking solution.

回溯

package com.shockang.study.algorithm.java.leetcode.leetcode101_200.leetcode131.solution1;import java.util.ArrayList;
import java.util.List;/*** 回溯** @author Shockang*/
public class Solution {public List<List<String>> partition(String s) {List<List<String>> res = new ArrayList<>();List<String> list = new ArrayList<>();dfs(s, 0, list, res);return res;}public void dfs(String s, int pos, List<String> list, List<List<String>> res) {if (pos == s.length()) res.add(new ArrayList<>(list));else {for (int i = pos; i < s.length(); i++) {if (isPal(s, pos, i)) {list.add(s.substring(pos, i + 1));dfs(s, i + 1, list, res);list.remove(list.size() - 1);}}}}public boolean isPal(String s, int low, int high) {while (low < high) if (s.charAt(low++) != s.charAt(high--)) return false;return true;}}

132. 分割回文串 II

Java Version, with comments and better naming. Easier to understand

动态规划

package com.shockang.study.algorithm.java.leetcode.leetcode101_200.leetcode132.solution1;import java.util.stream.IntStream;/*** @author Shockang*/
public class Solution {public int minCut(String s) {// validate inputif (s == null || s.length() <= 1) {return 0;}// dpint N = s.length();int[] dp = IntStream.range(0, N).toArray(); // initial value: dp[i] = ifor (int mid = 1; mid < N; mid++) { // iterate through all chars as mid point of palindrome// CASE 1. odd len: center is at index mid, expand on both sidesfor (int start = mid, end = mid; start >= 0 && end < N && s.charAt(start) == s.charAt(end); start--, end++) {int newCutAtEnd = (start == 0) ? 0 : dp[start - 1] + 1;dp[end] = Math.min(dp[end], newCutAtEnd);}// CASE 2: even len: center is between [mid-1,mid], expand on both sidesfor (int start = mid - 1, end = mid; start >= 0 && end < N && s.charAt(start) == s.charAt(end); start--, end++) {int newCutAtEnd = (start == 0) ? 0 : dp[start - 1] + 1;dp[end] = Math.min(dp[end], newCutAtEnd);}}return dp[N - 1];}
}

133. 克隆图

官方题解

Map+DFS

package com.shockang.study.algorithm.java.leetcode.leetcode101_200.leetcode133.solution1;import com.shockang.study.algorithm.java.leetcode.common.graph.Node;import java.util.HashMap;
import java.util.Map;
import java.util.stream.Collectors;/*** @author Shockang*/
public class Solution {private Map<Integer, Node> map = new HashMap<>();public Node cloneGraph(Node node) {if (node == null)return null;if (map.containsKey(node.val))return map.get(node.val);Node res = new Node(node.val);map.put(res.val, res);res.neighbors.addAll(node.neighbors.stream().map(this::cloneGraph).collect(Collectors.toList()));return res;}
}

HashMap,LinkedList,BFS

package com.shockang.study.algorithm.java.leetcode.leetcode101_200.leetcode133.solution2;import com.shockang.study.algorithm.java.leetcode.common.graph.Node;import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedList;/*** @author Shockang*/
public class Solution {public Node cloneGraph(Node node) {if (node == null) {return node;}HashMap<Node, Node> visited = new HashMap<>();// 将题目给定的节点添加到队列LinkedList<Node> queue = new LinkedList<>();queue.add(node);// 克隆第一个节点并存储到哈希表中visited.put(node, new Node(node.val, new ArrayList()));// 广度优先搜索while (!queue.isEmpty()) {// 取出队列的头节点Node n = queue.remove();// 遍历该节点的邻居for (Node neighbor : n.neighbors) {if (!visited.containsKey(neighbor)) {// 如果没有被访问过，就克隆并存储在哈希表中visited.put(neighbor, new Node(neighbor.val, new ArrayList()));// 将邻居节点加入队列中queue.add(neighbor);}// 更新当前节点的邻居列表visited.get(n).neighbors.add(visited.get(neighbor));}}return visited.get(node);}
}

134. 加油站

Share some of my ideas.

sumGas,sumCost,start,tank

package com.shockang.study.algorithm.java.leetcode.leetcode101_200.leetcode134.solution1;/*** @author Shockang*/
public class Solution {public int canCompleteCircuit(int[] gas, int[] cost) {int sumGas = 0;int sumCost = 0;int start = 0;int tank = 0;for (int i = 0; i < gas.length; i++) {sumGas += gas[i];sumCost += cost[i];tank += gas[i] - cost[i];if (tank < 0) {start = i + 1;tank = 0;}}if (sumGas < sumCost) {return -1;} else {return start;}}
}

135. 分发糖果

A simple solution

动态规划，初始都为 1，左一遍，右一遍

package com.shockang.study.algorithm.java.leetcode.leetcode101_200.leetcode135.solution1;import java.util.Arrays;/*** 动态规划，初始都为 1，左一遍，右一遍** @author Shockang*/
public class Solution {public int candy(int[] ratings) {int n = ratings.length;int[] candies = new int[n];Arrays.fill(candies, 1);for (int i = 1; i < n; i++) {if (ratings[i] > ratings[i - 1]) {candies[i] = candies[i - 1] + 1;}}int sum = candies[n - 1];for (int i = n - 2; i >= 0; i--) {if (ratings[i] > ratings[i + 1]) {candies[i] = Math.max(candies[i], candies[i + 1] + 1);}sum += candies[i];}return sum;}
}

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LeetCode 131~135

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