【力扣】每日三题 -- Day5

文章目录

• • • 41. 缺失的第一个正数
    • 209. 长度最小的子数组
    • 39. 组合总和 -- 回溯

41. 缺失的第一个正数

/*原地哈希*/
class Solution {
    public int firstMissingPositive(int[] nums) {
        for(int i = 0; i < nums.length; i++) {
            while(nums[i] > 0 && nums[i] <= nums.length && nums[i] != nums[nums[i] - 1] ) {
                swap(nums, i, nums[i]-1); //把nums[i]上的东西放到nums[nums[i]-1]的位置
            }
        }
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] != i + 1) {
                return i + 1;
            }
        }
        return nums.length + 1;
    }

    private void swap (int[] nums, int index1, int index2) {
        int temp = nums[index1];
        nums[index1] = nums[index2];
        nums[index2] = temp;
    }
}

209. 长度最小的子数组

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int left = 0, right = 0;
        int sum = 0;
        int minLen = Integer.MAX_VALUE;
        while(right < nums.length) {
            sum += nums[right];
            while(sum >= target) {
                minLen = Math.min(minLen, right - left + 1);
                sum -= nums[left];
                left ++; 
            }
            right ++;
        }
        return minLen = (minLen == Integer.MAX_VALUE) ? 0 : minLen;
    }
}

39. 组合总和 – 回溯

拿到这个题目能够想到是回溯+剪枝，但是和没思路差不多，没有想到要先排序。另外回溯“模板”还是不熟练，导致coding不上来。前几天做过的题和没做差不多。。。。。。
class Solution {
public List<List> combinationSum(int[] candidates, int target) {
List<List> res = new ArrayList<>();
List path = new ArrayList<>();
//if (candidates.length == 0) return res;
Arrays.sort(candidates);
backtrack(candidates, target, res, path, 0);
return res;
}
public void backtrack(int[] candidates, int target, List<List> res, List path, int begin) {
if(target < 0)
return;
if(target == 0){
res.add(new ArrayList<>(path));
return;
}
for(int i = begin; i < candidates.length; i ++) {
//剪枝
if(candidates[i] > target)
return;
if(i > 0 && candidates[i] == candidates[i-1])
continue;
//回溯套路
path.add(candidates[i]);
backtrack(candidates, target - candidates[i], res, path, i);
path.remove(path.size()-1);
}
}
}