第一思路,上来就暴力,直接两个for提交,搞定,注意>=target, 耗时106ms
class Solution {
public int minSubArrayLen(int target, int[] nums) {
int size = nums.length;
int minVal = Integer.MAX_VALUE;
for(int start = 0; start < size; start++){
int temp = target;
for(int end = start; end < size; end++){
temp -= nums[end];
if(temp <= 0){
int localSize = end - start + 1;
minVal = Math.min(minVal, localSize);
break;
}
}
}
return minVal == Integer.MAX_VALUE? 0 : minVal;
}
}
第二种思路、用一个滑动窗口,耗时1ms,学习此题,get到了滑窗的威力,把O(n**2)降到O(2n),每个元素进一次窗口,出一次窗口,涉及2次,共n个元素;
class Solution {
public int minSubArrayLen(int target, int[] nums) {
int size = nums.length;
int i = 0, j = 0;
int minVal = Integer.MAX_VALUE;
int sum = 0; //sum表示窗口内的和
for(j = 0; j < size; j++){
sum += nums[j];
while(sum >= target){
int subLength = j - i + 1;
minVal = Math.min(minVal, subLength);
sum -= nums[i++]; //移除起始位置
}
}
return minVal == Integer.MAX_VALUE? 0 : minVal;
}
}
重要学习资料,carl哥的代码随想录,牛掰!!
carlT209 长度最小子数组