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To solve the shortest path problem, of course, you need to think of the BFS: public int shortestPathBinaryMatrix(int[][] grid) { int[][] dirs = {{-1, 阅读全文
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Don't think about a lot, this is a very easy problem, time complexity is O(m*n) public boolean isToeplitzMatrix(int[][] matrix) { if (matrix == null | 阅读全文
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For the solution of this problem, we should be able to tell that it need a binary search. The left value is the maxmum weight in the weight list, beca 阅读全文
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My solution for this problem is using two stacks, it's very easy to understand: public boolean checkValidString(String s) { Stack<Integer> stars = new 阅读全文
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This problem is a typical backtracking solution problem, we need a recursive method as helper, the following is my first solution, which is not very g 阅读全文
该文被密码保护。 阅读全文
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Parentheses的题，首先想到stack，思路如下： 1. If get a left parentheses, if the stack size is odd number, that means, one right parentheses is missing, so result p 阅读全文
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这道题就是921的变种，话不多说： public int maxDepth(String s) { int count = 0; int max = 0; for (char c : s.toCharArray()) { if (c == '(') { count++; max = Math.max 阅读全文
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For this problem, if you can remember, always push the index of '(', the problem can be solved easily. The solution is: 1. if meet a '(', push the ind 阅读全文
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这道题是括号题，这种括号题如果是一种括号，就用一个int做stack就行，但是如果是多种括号，用int就烦琐了，一定要用Stack，下面是我的算法，时间复杂度就是O(n). 我把每类括号的做括号和右括号放在map中，如果是左括号就push到stack，如果是右括号，就pop stack，map中对应 阅读全文