leetcode 377 Combination Sum IV 动态规划

377. Combination Sum IV

 

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• Total Accepted: 3456
• Total Submissions: 8922
• Difficulty: Medium

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]target = 4The possible combination ways are:(1, 1, 1, 1)(1, 1, 2)(1, 2, 1)(1, 3)(2, 1, 1)(2, 2)(3, 1)Note that different sequences are counted as different combinations.Therefore the output is 7.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

/** *这个方法每次重复计算已经算过的内容，会超时  * public class Solution {    public int combinationSum4(int[] nums, int target) {        //Arrays.sort(nums);        int len=nums.length;        //int flag = Arrays.BinarySearch(nums,target);        int flag=-1;        int ret=0;        if(len==1&&nums[0]!=target)return 0;        if(len==1&&nums[0]==target)return 1;        for(int i=0;i<len;i++){            if(nums[i]==target){                flag=i;                break;            }        }        if(flag>0){            ret=1+combinationSum4(nums,target-nums[flag]);        }        if(flag<0){            int i;            for(i=len-1;i>=0;i--){                if(nums[i]<target){                    break;                }            }            int[] temp=new int[i+1];            for(int j=0;j<i;j++)temp[j]=nums[j];            ret=combinationSum4(nums,target-nums[i])+combinationSum4(temp,target);                    }        return ret;    }} *//** *这个方法每次计算记录下已经算过的内容，不用重复，更简单 * 这个是参考的。。。 *  */ public class Solution {    public int combinationSum4(int[] nums, int target) {        if(nums == null || nums.length == 0 || target < 0) {return 0;    }    Arrays.sort(nums); //Not required, but can help in exit out of loop early.    int[] combinations = new int[target + 1];    combinations[0] = 1;    int length = nums.length; // so that length is not calculated every time.    for (int i = 1; i <= target; i++) {    for (int j = 0; j < length; j++) {    if(i - nums[j] < 0) {    break;    }    if( i == nums[j]) {    combinations[i] += 1;    }    else if (i - nums[j] >= 0) {    combinations[i] += combinations[i - nums[j]];    }    }    }    return combinations[target];    } }