给定一个整数数组和一个整数，返回两个数组的索引

public static int[] twoSum(int[] numbers, int target) { int i = 0, j = numbers.length - 1;

    while (i != j) {
        if (numbers[i] + numbers[j] == target) {
            return new int[]{i + 1, j + 1};
        }

        if (numbers[i] + numbers[j] < target) {
            i++;
            continue;
        }

        if (numbers[i] + numbers[j] > target) {
            j--;
            continue;
        }
    }

    return new int[]{i, j};
}

参考答案：

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

分析：空间复杂度和时间复杂度均为 O(n)