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].
解法一：
暴力解决很简单,但时间复杂度为O(n^2)。
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] result=new int[2];
for(int i=0;i<nums.length-1;i++){
for(int j=i+1;j<nums.length;j++){
if(nums[i]+nums[j]==target){
return new int[]{i,j};
}
}
}
return result;
}
}
解法二：
先遍历一遍数组，建立map数据，然后再遍历一遍，开始查找，找到则记录index。时间复杂度为O(n)。
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
m.put(nums[i], i);
}
for (int i = 0; i < nums.length; ++i) {
int t = target - nums[i];
if (m.containsKey(t) && m.get(t) != i) {
res[0] = i;
res[1] = m.get(t);
break;
}
}
return res;
}
}