1. Two Sum

Question

1. Two Sum

Solution

思路很简单这里就不说了，下面贴了不同的几个Java实现的时间与其他算法实现的时间的比较

这个是LeetCode的第一道题，也是我刷的第一道，刚一开始的Java实现

public int[] twoSum(int[] nums, int target) {
    boolean find = false;
    int targeti = -1, targetj = -1;
    for (int i = 0; i < nums.length; i++) {
        for (int j = 0; j < nums.length; j++) {
            if (i == j) {
                continue;
            }
            if (nums[i] + nums[j] == target) {
                targeti = i;
                targetj = j;
                find = true;
                break;
            }
        }
        if (find) {
            break;
        }
    }
    return new int[]{targeti, targetj};
}

稍微改进一下

public int[] twoSum(int[] nums, int target) {
    boolean find = false;
    int targeti = -1, targetj = -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) {
                targeti = i;
                targetj = j;
                find = true;
                break;
            }
        }
        if (find) {
            break;
        }
    }
    return new int[]{targeti, targetj};
}

下面这个版本是在讨论中看到的O(1)时间复杂度

public int[] twoSum3(int[] nums, int target) {
    int[] result = new int[2];
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int i = 0; i < nums.length; i++) {
        if (map.containsKey(target - nums[i])) {
            result[1] = i;
            result[0] = map.get(target - nums[i]);
            return result;
        }
        map.put(nums[i], i);
    }
    return result;
}