2017-03-08 01:01:31 +0000   |     algorithm leetcode array hash table   |   Viewed times   |    

题目

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].

log(n2)朴素方法

最简单的两层迭代。

public static int[] twoSum(int[] nums, int target) {
    if (nums.length < 2) {
        throw new IllegalArgumentException();
    }
    int[] result = new int[] {-1,-1};
    for (int i = 0; i < nums.length-1; 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 result;
}

结果

two-sum-1

log(n)解法

空间换时间。用Map存下以前遇到过的值。

public static int[] twoSumLogN(int[] nums, int target) {
    if (nums.length < 2) {
        throw new IllegalArgumentException();
    }
    Map<Integer,Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int diff = target - nums[i];
        if (map.containsKey(diff)) {
            return new int[] {map.get(diff),i};
        }
        map.put(nums[i],i);
    }
    return new int[] {0,0};
}

结果

two-sum-2