[LeetCode]Two Sum
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解题思路:
1,暴力法,O(n^2) time O(1) space
2,可以利用 hash表,把 数组中的值 映射到 index, 每遍历到一个数x,就看一下hash表中是否存在 target-x,如果存在, 就return了;不存在,就把x加入到 hash表中
时间复杂度 O(n),空间复杂度O(n)
public class Solution { public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; ++i){ int x = nums[i]; if (map.containsKey(target-x)){ return new int[] {map.get(target-x) + 1, i + 1}; } map.put(x, i); } throw new IllegalArgumentException("no to sum solution"); }}
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