Getting Started with LeetCode Algorithm - Input Sorted Array

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Sum of Two Numbers II - Input Sorted Array

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Given an integer array  with subscripts starting from 1numbers , the array has been arranged in non- decreasing order  , please find two numbers from the array that satisfy the addition and the sum is equal to the target  targetnumber. If these two numbers are respectively numbers[index1] and numbers[index2], then 1 <= index1 < index2 <= numbers.length.

Returns the sum of the subscripts of these two integers as an array of of length 2 .[index1, index2]index1index2

You can assume that each corresponds to a unique answer , and you cannot reuse the same elements.

The solution you design must use only constant-level extra space. Example 1:

输入：numbers = [2,7,11,15], target = 9
输出：[1,2]
解释：2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。
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Example 2:

输入：numbers = [2,3,4], target = 6
输出：[1,3]
解释：2 与 4 之和等于目标数 6 。因此 index1 = 1, index2 = 3 。返回 [1, 3] 。
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Example 3:

输入：numbers = [-1,0], target = -1
输出：[1,2]
解释：-1 与 0 之和等于目标数 -1 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。
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Thought analysis

method one

Simple and rude double-layer loop, judge the condition, and then return to the subscript.

Method Two

1. Because the array is ordered, binary search can be used.
2. To find the sum of two numbers equal to the target value, you can fix one number first, then find another number, and the other number is equal to the target value minus the first number.
3. The left boundary, right boundary and median are still defined, and the circular movement can be obtained.

AC code

method one

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    const len = numbers.length
    for(let i = 0; i < len; i++) {
        for(let j = i + 1; j < len; j++) {
            if(numbers[i] + numbers[j] === target) {
                return [i + 1, j + 1]
            }
        }
    }
};
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result:

• Execution result: passed
• Execution time: 548 ms, beating 5.06% of users across all JavaScript commits
• Memory consumption: 42 MB, beats 44.43% of users across all JavaScript submissions
• Test cases passed: 21/21

Method Two

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    for (let i = 0; i < numbers.length; i++) {
        let low = i + 1
        let high = numbers.length - 1
        while (low <= high) {
            let mid = Math.floor((high - low) / 2) + low;
            if (numbers[mid] === target - numbers[i]) {
                return [i + 1, mid + 1];
            } else if (numbers[mid] > target - numbers[i]) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
    }
};
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result:

• Execution result: passed
• Execution time: 64 ms, beating 81.18% of users across all JavaScript submissions
• Memory consumption: 42.3 MB, beats 17.11% of users across all JavaScript submissions
• Test cases passed: 21/21

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