两数之和 Two Sum II - Input array is sorted

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问题：

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

解决：

①暴力方法，从头到尾扫描找到满足条件的两个值，超时O（N^2）。

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = new int[2];
        for (int i = 0;i < numbers.length ;i ++ ) {
            for (int j = i + 1;j < numbers.length ;j ++ ) {
                if (numbers[i] + numbers[j] == target) {
                    res[0] = i + 1;
                    res[1] = j + 1;
                }
            }
        }
        return res;
    }
}

②使用Map，存储nums[i],i，当可以在map中找到target - nums[i]时，返回i+1和target - nums[i]的下标+1。耗时5ms.

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = new int[2];
        Map<Integer,Integer> map = new HashMap<>();
        for (int i = 0;i < numbers.length ;i ++ ) {
            if(map.containsKey(target - numbers[i])){
                res[0] = map.get(target - numbers[i]) + 1;
                res[1] = i + 1;
                break;
            }
            map.put(numbers[i],i);
        }
        return res;
    }
}

③使用双指针+二分查找（排好序）。首先令start = 0，end = numbers.length-1。比较：

···若numbers[start] + numbers[end] < target ， 则 start 不动，使用二分查找法在 [start+1，end-1]中确定新的end值，比较对象为target - numbers[start] ；

···若numbers[start] + numbers[end] > arget ， 则 end 不动，使用二分查找法在 [start+1，end-1]中确定新的start值，比较对象为target - numbers[end] ；耗时0ms。

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int start = 0,end = numbers.length - 1;
        int[] res = new int[2];
        while (start < end) {
            int sum = numbers[start] + numbers[end];
            if (sum == target) {
                res[0] = start+1;
                res[1] = end+1;
                return res;
            } else if (sum < target) {
                start = binarySearch(numbers, start + 1, end - 1, target - numbers[end]);
            } else {
                end = binarySearch(numbers, start + 1, end - 1, target - numbers[start]);
            }
        }
        return res;
    }

    private int binarySearch(int[] nums, int start, int end, int target) {
        while (start < end) { //注意是<，而非一般二分查找的 <=，它可能会导致无限循环
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return start;
    }
}

④双指针（排好序），一个从左边开始，一个从右边开始，如果和比target小，则左边指针右移，如果和比target大，则右边指针左移，直到和等于target为止。耗时2ms。

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = new int[2];
        int start = 0;
        int end = numbers.length - 1;
        while(start < end){
            int sum = numbers[start] + numbers[end];
            if (sum == target) {
                res[0] = start + 1;
                res[1] = end + 1;
                break;//一定要有break,否则超时
            }else if(sum < target){
                start ++;
            }else if(sum > target){
                end --;
            }
        }
        return res;
    }
}

转载于:https://my.oschina.net/liyurong/blog/915794