leetcode 1. Two Sum -----java
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题目链接:https://leetcode.com/problems/two-sum/
题目:
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.
Example:
Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].
这道题一看就知道用暴力搜索肯定没问题,而且猜到OJ肯定不会允许用暴力搜索这么简单的方法,于是去试了一下,果然是Time Limit Exceeded,这个算法的时间复杂度是O(n^2)。
class Solution {
public int[] twoSum(int[] nums, int target) {
int [] res = new int[2];
for(int i=0;i<nums.length;i++){
for(int j=i+1;j<nums.length;j++){ //注意这里i+1
if(nums[i]+nums[j]==target){
res[0] = i;
res[1] = j;
}
}
}
return res;
}
}
解析:利用hashmap,key存放数值,value存放出现的位置。从前到后进行遍历,将target值减去当前的值,看是否存在map中,
若存在map中则取出相应的标号,退出。
java代码实现:
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int result[] = new int[2];
for (int i = 0; i < nums.length; i++){
int num = target - nums[i];
if(map.containsKey(num)){
result[0] = map.get(num);
result[1] = i;
return result;
}
map.put(nums[i], i);
}
return result;
}
}
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