Leetcode 之路 - 1. Two Sum

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1. Two Sum

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Description

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].

Thinking

• 返回数组 nums 中两个值的下标，这两个值的和等于所给的 target
• 数组 nums 中的元素并不是有序的
• 假定解是唯一的，并且每个元素不得重复使用
• 需要特别注意元素是 target 的一半的情况
• 还需要特别注意一个值重复出现的情况

Solution -1

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

    Map<Integer, Integer> map = new HashMap<>();

    // 保存每个值到 map 中，value 取用同 target 的差
    for (int i = 0; i < nums.length; i++) {
        if ((target-nums[i]) == nums[i] && !map.containsKey(nums[i])) {
            map.put(nums[i], -1);
        } else {
            map.put(nums[i], target-nums[i]);
        }
    }

    int[] result = new int[2];

    for (int i = 0; i < nums.length; i++) {
        if (map.containsKey(map.get(nums[i]))) {
            result[0] = i;
            break;
        }
    }

    for (int i = 0; i < nums.length; i++) {
        if (i != result[0] && nums[i] == map.get(nums[result[0]])) {
            result[1] = i;
            break;
        }
    }

    return result;

}

Solution -2

大神做法，借鉴学习

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

    Map<Integer, Integer> res = new HashMap<>();

    for (int i = 0; i < nums.length; i++) {
        int difference = target - nums[i];
        if (res.containsKey(difference)) {
            return new int[]{ res.get(difference), i };
        }
        res.put(nums[i], i);
    }
    return null;

}