【leetcode】16. 最接近的三数之和

• 三重循环
• 排序+双指针
• • 判断重复
  • 不判断重复
• 总结

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        length = len(nums)
        minsum = float("inf")
        p1 = 0
        while p1 < length:
            p2 = p1 + 1
            while p2 < length:
                p3 = p2 + 1
                while p3 < length:
                    if abs(minsum) > abs(target - (nums[p1] + nums[p2] + nums[p3])):
                        minsum = target - (nums[p1] + nums[p2] + nums[p3])
                        best = nums[p1] + nums[p2] + nums[p3]
                    p3 += 1
                p2 += 1
            p1 += 1
        return best

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        length = len(nums)
        minsum = float("inf")
        best = 0
        for i in range(length):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            L = i + 1
            R = length - 1
            while L < R:
                while L < R and nums[L] == nums[L + 1]:
                    L += 1
                while L < R and nums[R] == nums[R - 1]:
                    R -= 1
                sum = target - (nums[i] + nums[L] + nums[R])
                if minsum > abs(sum):
                    minsum = abs(sum)
                    best = nums[i] + nums[L] + nums[R]
                if sum == 0:
                    return best
                elif sum > 0:
                    L += 1
                else:
                    R -= 1
        return best

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        length = len(nums)
        sumdifference = float("inf")
        last = 0
        for i in range(length):
            L = i + 1
            R = length - 1
            while L < R:
                sum = target - (nums[i] + nums[L] + nums[R])
                if sumdifference > abs(sum):
                    sumdifference = abs(sum)
                    last = nums[i] + nums[L] + nums[R]
                if sum == 0:
                    return last
                elif sum > 0:
                    L += 1
                else:
                    R -= 1
        return last

1、思路同15题，但是由于只需输出最接近target的元素和，所以可以不进行重复数组判断。
2、暴力破解的三重循环，不出所料依旧超时。
3、python中的无穷大表示方法num = float(“inf”)。