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].
这道题要求返回两个数值i和j，使num[i]+num[j]=target;
这道题一开始的想法就是两个for循环，即
public class Solution{
public int[] twoSum(int[] nums,int target){
int[] back = 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){
back[0]=i;
back[1]=j;
return back;
}
}
}
return null;
}
}
然后在网上看到了另一种时间复杂度更低的算法，贴上来也看一下：
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] back = new int[2];
Map map = new HashMap<>();
for(int i=0;i<nums.length;i++){
map.put(nums[i], i);
}
for(int i=0;i<nums.length;i++){
int delNum = target-nums[i];
if(map.get(delNum)!=null&&(int)map.get(delNum)>i){
back[0]=i;
back[1]=(int)map.get(delNum);
}
}
return back;
}
}
