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].

---

方案1：时间复杂度（O(n^2)）

class Solution {

public int[] twoSum(int[] nums, int target) {

int[] result = new int[2];

for(int i = 0; i < nums.length; i++) {

for(int j = i+1; j < nums.length; j++) {

if(nums[i] + nums[j] == target) {

result[0] = i;

result[1] = j;

return result;

}

}

}

return null;

}

}

优化版本：时间复杂度（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++){
            Integer index=map.get(target-nums[i]);
            if(index==null){
                map.put(nums[i],i);
            }else{
                return new int[]{i,index};
            }
        }
        return new int[]{0,0};
    }
}

方案2来自： https://blog.csdn.net/lnho2015/article/details/50889558