LeetCode: 1. Two Sum

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1. Two Sum

LeetCode - 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].

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/**
 * 原始算法，时间复杂度为 `O(n^2^)`.
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2018-06-30
 */
public static int[] twoSum(int[] nums, int target) {
    if (nums == null || nums.length < 2) {
        return null;
    }
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] == target) {
                return new int[]{i, j};
            }
        }
    }

    return null;
}

/**
 * 优化后的算法，时间复杂度减低为 `O(n)`， 空间复杂度提高了。
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2018-07-13 23:31
 */
public static int[] twoSumO1(int[] nums, int target) {
    if (nums == null || nums.length < 2) {
        return null;
    }

    Map<Integer, Integer> diffNumToIndexMap = new HashMap<>(nums.length * 4 / 3);
    for (int i = 0; i < nums.length; i++) {
        if (diffNumToIndexMap.containsKey(nums[i])) {
            return new int[]{diffNumToIndexMap.get(nums[i]), i};
        } else {
            diffNumToIndexMap.put(target - nums[i], i);
        }
    }

    return null;
}