回溯

全排列

class Solution {/**回溯问题套路模板bactracing(nums, startIndex)*/List<List<Integer>> res = new ArrayList();List<Integer> path = new ArrayList();public List<List<Integer>> permute(int[] nums) {if(nums.length == 0 || nums == null) {return res;}backtracing(nums, 0);return res;}public void backtracing(int[] nums, int startIndex) {if(path.size() == nums.length) {res.add(new ArrayList(path));return;}for(int i = 0; i < nums.length; i++) {if(!path.contains(nums[i])) {path.add(nums[i]);backtracing(nums, i + 1);path.remove(path.size() - 1);}       }}
}

子集

class Solution {/**和上一题的区别就是，少加一步判断for循环从startIndex开始*/List<List<Integer>> res = new ArrayList();List<Integer> path = new ArrayList();public List<List<Integer>> subsets(int[] nums) {if(nums.length == 0 || nums == null) {return res;}Arrays.sort(nums);backtracing(nums, 0);return res;}public void backtracing(int[] nums, int startIndex) {if(!res.contains(path)) {res.add(new ArrayList(path));}for(int i = startIndex; i < nums.length; i++) {if(!path.contains(nums[i])) {path.add(nums[i]);backtracing(nums, i + 1);path.remove(path.size() - 1);}}}
}

电话号码的字母组合

class Solution {/**先创建一个数字和字母对应的hashmap回溯的时候，遍历两个字符串*/List<String> res = new ArrayList();StringBuffer path = new StringBuffer();HashMap<Character, String> map = new HashMap();public List<String> letterCombinations(String digits) {map.put('2', "abc");map.put('3', "def");map.put('4', "ghi");map.put('5', "jkl");map.put('6', "mno");map.put('7', "pqrs");map.put('8', "tuv");map.put('9', "wxyz");if(digits == null ||digits.length() == 0) {return res;}backtracing(digits, 0);return res;}public void backtracing(String digits, int startIndex) {if(path.length() == digits.length()) {res.add(path.toString());return;}char digit = digits.charAt(startIndex);String s = map.get(digit);char[] ch = s.toCharArray();for(char c: ch) {path.append(c);backtracing(digits, startIndex + 1);path.deleteCharAt(path.length() - 1);}}
}

组合总和

class Solution {/**可以被重复选取*/List<List<Integer>> res = new ArrayList();List<Integer> path = new ArrayList();public List<List<Integer>> combinationSum(int[] candidates, int target) {backtracing(candidates, target, 0);return res;}public void backtracing(int[] candidates, int target, int start) {if(sum(path) == target) {res.add(new ArrayList(path));return;}else if(sum(path) > target) {return;}// #############################################// 其他方式?for(int i = start; i < candidates.length; i++) {path.add(candidates[i]);backtracing(candidates, target, i);path.remove(path.size() - 1);}}public int sum(List<Integer> path) {int sum = 0;for(int i = 0; i < path.size(); i++) {sum += path.get(i);}return sum;}
}

括号生成

class Solution {/**left和right分别表示左括号和右括号数量*/List<String> res = new ArrayList();StringBuffer path = new StringBuffer();public List<String> generateParenthesis(int n) {backtracing(n, 0, 0);return res;}public void backtracing(int n, int left, int right) {if(path.length() == n * 2) {res.add(path.toString());return;}if(left < n) {path.append("(");backtracing(n, left + 1, right);path.deleteCharAt(path.length() - 1);}if(right < left) {path.append(")");backtracing(n, left, right + 1);path.deleteCharAt(path.length() - 1);}}
}

单词搜索

分割回文串

贪心算法

买卖股票都最佳时期

class Solution {/**0      1dp[i]表示第i天买入 or 卖出买入 dp[i][0] = dp[i - 1][0], dp[i - 1][1] + prices[i];dp[i][1] = dp[i - 1][1], -prices[i]*/public int maxProfit(int[] prices) {int n = prices.length;int[][] dp = new int[n][2];dp[0][1] = -prices[0];for(int i = 1; i < n; i++) {dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);dp[i][1] = Math.max(dp[i - 1][1], -prices[i]);}return dp[n - 1][0];}
}

跳跃游戏

class Solution {/**朴实无华的想法不太对boolean dp[i]表示可以第i个位置是可以到达的dp[i] = dp[i - 1] ? dp[i - 1] >= 1，dp[i] = true ?##################################################dp[i]表示第i个位置可以跳跃的最远距离？dp[i] i + nums[i]*/public boolean canJump(int[] nums) {int[] dp = new int[nums.length];if(nums.length == 0 || nums == null) {return false;}dp[0] = nums[0];// 0 1 2 3 4// 2,3,1,1,4// 2,4,4,4,8// 0 1 2// 0 2 2// 0 xfor(int i = 1; i < dp.length; i++) {if(dp[i - 1] < i) {return false;}dp[i] = Math.max(dp[i - 1], i + nums[i]);if(dp[i] >= nums.length - 1) {return true;}}return dp[dp.length - 1] >= nums.length - 1;}
}

跳跃游戏II

class Solution {/**dp[i]表示跳跃到位置i的最小跳跃次数感觉像上一题那样直接求出dp[i]然后看变化了几次？*/public int jump(int[] nums) {int jump = 0;int maxdistance = 0;int index = 0;for(int i = 0; i < nums.length - 1; i++) {maxdistance = Math.max(i + nums[i], maxdistance);if(index == i) {jump++;index = maxdistance;}}return jump;}
}

划分字母区间

class Solution {/**记录s中所有字母出现的最后一个位置用int[26]记录每次遍历更新endend - start + 1*/public List<Integer> partitionLabels(String s) {int[] lastIndex = new int[26];List<Integer> list = new ArrayList();for(int i = 0; i < s.length(); i++) {lastIndex[s.charAt(i) - 'a'] = i;}int start = 0;int end = 0;for(int i = 0; i < s.length(); i++) {end = Math.max(end, lastIndex[s.charAt(i) - 'a']);if(end == i) {list.add(end - start + 1);start = end + 1;}}return list;}
}