leetcode-two sum
2017-05-09 11:13
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lz决定增长自己的算法水平了,因为自己不可能一直做低层次的编写代码工作。lz最近在关注牛客网,他真的是一个很棒的网站,在这里更加看清了自己和外面的差距,同时也找到了解决差距的办法。其中就有leetcode,好了闲话少叙,开始记录我的LeetCode!
1. Two Sum:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
代码:
/*** * leetcode 1.two sum * 题意:给定一个整数数组nums和一个整数target,要你求出nums数组哪两个元素相加等于target,并返回两个元素的下标 * 约束条件:1.两个元素的下标不能相同,即不能是同一个元素 2.nums总能有正解 * 解题思路:既然两个元素不能相同,则用map来保存数组元素和下标,键为数组元素,值为数组下标。循环遍历nums,将当前数组元素放入到map中,判断如果map中含有target-nums[i]的键,有则nums[i]就是第二个元素,target-nums[i]就是第一个元素 * @param nums * @param target * @return */
public int[] twoSum(int[] nums, int target) { int[] result = new int[2]; Map<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(target - nums[i])) { result[1] = i; result[0] = map.get(target - nums[i]); return result; } map.put(nums[i], i); } return result; }
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