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LeetCode1-Two Sum

2016-07-25 23:46 309 查看
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

题目:给定一数组,输出满足条件的数组下标,实现的思想比较简单,先实现排序,然后采用start、end下标遍历满足条件的下标。

/*
* time:2016-7-25 22:08:24
* location:home
* */

import java.util.Arrays;

public class Solution {
public static int[] twoSum(int[] nums, int target){

//		int[] res = new int[2];
//		res[0] = -1;
//		res[1] = -1;
//
//		final HashMap<Integer, Integer> h = new HashMap<Integer,Integer>();
//
//		for(int i = 0; i < nums.length; i++){
//			if(h.containsKey(target - nums[i])){
//				int index = h.get(target-nums[i])+1;
//				res[0] = Math.min(i+1, index);
//				res[1] = Math.max(i+1, index);
//			}
//			h.put(nums[i],i);
//		}
//
//		return res;

//O(nlogn)
int a = 0, b=0;
int start = 0, end = nums.length-1;
int [] nums2 = Arrays.copyOf(nums, nums.length);
Arrays.sort(nums2);
//find the nums,头尾指针
while(start < end){
int sum = nums2[start] + nums2[end];
if(sum < target){
start++;
}
else if(sum > target){
end--;
}
else{
a = nums2[start]; b = nums2[end];
break;
}
}
//find the index of the nums
int[] res = new int[2];
for(int i = 0; i < nums.length; i++){
if(nums[i] == a){
res[0] = i;
break;
}
}

if(a != b){
for(int i =0; i < nums.length; i++){
if(nums[i] == b){
res[1] = i;
break;
}
}
} else {
for(int i = 0; i < nums.length; i++){
if(nums[i] ==b && i != res[0]){
res[1] = i;
break;
}
}
}
return res;

}
//test
public static void main(String[] args){
int[] nums = {3,2,2,7,7,8,11,15};
int tagert = 9;

int[] result = twoSum(nums,tagert);

for(int i=0; i < result.length; i++){
System.out.println("res[i]"+result[i]);
}

}
//	public int[] TwoSum(int[] nums, int target){
//
//	}

}
代码参考leetcode discussion
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