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  • 3 Sum Closest

    问题描述

    Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

        For example, given array S = {-1 2 1 -4}, and target = 1.
    
        The sum that is closest to the target is 2. (-1 + 2 + 1 = 2)

    解决思路

    1. 搜索(超时);

    2. 排序 + 二分, 时间复杂度为O(n^2).

    程序

    public class ThreeSumClosest {
    	// dfs
    	private int diff = 0;
    	private int abs = Integer.MAX_VALUE;
    
    	public int threeSumClosest(int[] nums, int target) {
    		if (nums == null || nums.length < 3) {
    			return -1;
    		}
    
    		List<Integer> sol = new ArrayList<Integer>();
    		boolean[] used = new boolean[nums.length];
    		helper(nums, target, sol, used, 0);
    
    		return target - diff;
    	}
    
    	private void helper(int[] nums, int target, List<Integer> sol,
    			boolean[] used, int begin) {
    		if (sol.size() == 3) {
    			if (Math.abs(target) < abs) {
    				abs = Math.abs(target);
    				diff = target;
    			}
    			return;
    		}
    		for (int i = begin; i < nums.length; i++) {
    			if (used[i]) {
    				continue;
    			}
    			used[i] = true;
    			sol.add(nums[i]);
    			helper(nums, target - nums[i], sol, used, begin + 1);
    			used[i] = false;
    			sol.remove(sol.size() - 1);
    		}
    	}
    	
    	// sort + bs
    	public int threeSumClosest(int[] nums, int target) {
    		if (nums == null || nums.length < 3) {
    			return -1;
    		}
    
    		Arrays.sort(nums);
    		int diff = 0;
    		int abs = Integer.MAX_VALUE;
    
    		for (int i = 0; i < nums.length - 2; i++) {
    			int begin = i + 1;
    			int end = nums.length - 1;
    			while (begin < end) {
    				int sum = nums[i] + nums[begin] + nums[end];
    				if (sum == target) {
    					return sum;
    				}
    				if (sum < target) {
    					++begin;
    				} else {
    					--end;
    				}
    				if (Math.abs(sum - target) < abs) {
    					abs = Math.abs(sum - target);
    					diff = sum - target;
    				}
    			}
    		}
    
    		return target + diff;
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/harrygogo/p/4675411.html
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