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].
最笨的方法:
public class Solution {
public int[] twoSum(int[] nums, int target) {
int [] res={1,2};
for(int i=0;i<nums.length;i++){
int temp=nums[i];
for(int j=i+1;j<nums.length;j++){
if(temp+nums[j]==target){
res[0]=i; res[1]=j;
}
}
}
return res;
}
}
较好的是使用hash:
public class Solution {
public int[] twoSum(int[] nums, int target) {
int [] res=new int[2];
Map<Integer, Integer> m = new HashMap<>();
for(int i=0;i<nums.length;i++){
if(m.get(target-nums[i])!=null){
res[0]=i;
res[1]=m.get(target-nums[i]);
return res;
}
m.put(nums[i],i);
}
return res;
}
}
本文介绍了一种经典的算法问题——寻找数组中两个数相加等于特定目标值的索引。给出了两种解决方案:一种是最笨的方法,通过双重循环实现;另一种是更高效地利用哈希表来解决该问题。
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