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The idea is to search for the left and right boundaries of target via two binary searches. Well, some tricks may be needed. Take a look at this link :... 阅读全文
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The most obvious idea is to maintain two divisors to get the most and least significantdigits and compare them. Well, there are much more clever ideas... 阅读全文
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The idea is to add the two numbers (represented by linked lists) nodes after nodes and store the result in the longer list. Since we may append the ad... 阅读全文
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The basic idea is to maintain a hash table for each elementnuminnums, usingnumas key and its index (1-based) as value. For eachnum, search fortarget -... 阅读全文
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This problem can be solved easily if we are allowed to use more than O(1) space. For example, you may create a copy of the original matrix (O(mn)-spac... 阅读全文
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This problem is a little tricky at first glance. However, if you have finished theHouse Robberproblem, this problem can simply bedecomposed into two H... 阅读全文
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Since we are not allowed to rob two adjacent houses, we keep two variablespreandcur. During thei-th loop,prerecords the maximum profit that we do not ... 阅读全文
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Well, this problem can be solved in 1-line clearly. Take a look at this link:-)1 class Solution {2 public:3 string convertToTitle(int n) {4 ... 阅读全文