zoukankan      html  css  js  c++  java
  • LeetCode 16.3Sum Closest

    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).

     1 class Solution4 {
     2 //    时间复杂度O(n3)
     3 //    public int threeSumClosest(int[] nums, int target) {
     4 //        Arrays.sort(nums);
     5 //        int closest = 65535, result = -65535;
     6 //        for (int i = 0; i < nums.length - 2; i++) {
     7 //            for (int low = i+1; low < nums.length - 1; low++) {
     8 //                for (int high = nums.length - 1; high > low; high--) {
     9 //                    int sum = nums[i] + nums[low] + nums[high];
    10 //                    if (closest > Math.abs(target - sum)){
    11 //                        closest = Math.abs(target - sum);
    12 //                        result = sum;
    13 //                    }
    14 //                }
    15 //            }
    16 //        }
    17 //        return result;
    18 //    }
    19 
    20 //    时间复杂度O(n3) 前进了一小步
    21 //    public int threeSumClosest(int[] nums, int target) {
    22 //        Arrays.sort(nums);
    23 //        int closest = 65535, result = -65535;
    24 //        for (int i = 1; i < nums.length - 1; i++) {
    25 //            for (int low = i-1; low >= 0; low--) {
    26 //                for (int high = i+1; high < nums.length; high++) {
    27 //                    int sum = nums[low] + nums[i] + nums[high];
    28 //                    if (closest > Math.abs(target - sum)){
    29 //                        closest = Math.abs(target - sum);
    30 //                        result = sum;
    31 //                    }
    32 //                }
    33 //            }
    34 //        }
    35 //        return result;
    36 //    }
    37 
    38 //  前进了一大步
    39     /*
    40     可以看出while与for的区别:
    41         while更灵活,能够根据情况控制step;for则不行
    42      */
    43     public int threeSumClosest(int[] nums, int target) {
    44         Arrays.sort(nums);
    45         int closest = 65535, result = -65535;
    46         for (int i = 0; i < nums.length - 2; i++) {
    47             int low = i + 1, high = nums.length - 1;
    48             while (low < high){
    49                 int sum = nums[i] + nums[low] + nums[high];
    50                 if (closest >Math.abs(target-sum)){
    51                     closest = Math.abs(target-sum);
    52                     result = sum;
    53                 }
    54                 if (sum > target){
    55                     --high;
    56                 }else if(sum < target){
    57                     ++low;
    58                 }else {//正好相等
    59                     return sum;
    60                 }
    61             }
    62         }
    63         return result;
    64     }
    65 }
  • 相关阅读:
    react脚手架和JSX
    promise
    防抖和节流
    call/apply/bind 用法
    js this指向
    vue单页面应用刷新网页后vuex的state数据丢失的解决方案
    Echarts基础
    继承
    原型链
    vue项目中使用生成动态二维码
  • 原文地址:https://www.cnblogs.com/yfs123456/p/10889846.html
Copyright © 2011-2022 走看看