LeetCode之3Sum Closest（Kotlin）

问题： Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
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方法： 最原始的方法是三层循环遍历，求全部三个数的sum，而后找出最接近target的sum，可是这样的算法复杂度比较高，须要O(n^3)的时间复杂度，经过排序能够下降比较的次数，先对数组进行排序，由于数组有序，若是sum大于target就从后面移动index，若是sum小于target就从前面移动index，向中间逼近，这样就能够减小比较的次数，这样的算法时间复杂度是O(n^2)

具体实现：

class ThreeSumClosest {
    // -4, -1, 2, 1,
    // 双逼近算法
    fun threeSumClosest(nums: IntArray, target: Int): Int {
        var sum = nums[0] + nums[1] + nums[2]
        nums.sort()
        for (x in nums.indices) {
            var i = x + 1
            var j = nums.lastIndex
            while (i < j) {
                val curSum = nums[i] + nums[j] + nums[x]
                val curDiff = abs(sum - target)
                val diff  = abs(curSum - target)
                if (diff < curDiff) {
                    sum = curSum
                }
                if (curSum - target > 0) {
                    j--
                } else if(curSum - target < 0) {
                    i++
                } else {
                    return sum
                }
            }
        }
        return sum
    }
}

fun main(args: Array<String>) {
    val nums = intArrayOf(0,2,1,-3)
    val target = 1
    val threeSumClosest = ThreeSumClosest()
    println(threeSumClosest.threeSumClosest(nums, target))
}
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有问题随时沟通

具体代码实现能够参考Github