LeetCode 209. Minimum Size Subarray Sum

问题描述

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,

the subarray [4,3] has the minimal length under the problem constraint.

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Credits:

Special thanks to @Freezen for adding this problem and creating all test cases.

问题分析

给定一个数组nums[],给定一个数target。求nums中连续子集的和等于target。求出这些子集中长度最小的。遇到连续子集算法的实现有两种，一种是滑动窗口，窗口的条件是sum>= target,这种的复杂度是O(n)。第二种是求前n个的和，这里需要把sum[0]设置为0，这个算法的复杂度是O(nlogn)

代码实现

public int minSubArrayLen(int s, int[] nums) {

//找出和大于s的 nums中的子序列最小的长度
//使用滑动窗口中和》= nums的的窗口,找到

int minLength = Integer.MAX_VALUE;
int leftIndex = 0;
int rightIndex = 0;
int sum = 0;
int length = 0;
while (sum >= s || rightIndex < nums.length) {//如何确定边界

if (sum < s) {// 在小于的时候
length++;
sum = sum + nums[rightIndex];
rightIndex++;
} else {//大于等于的时候
minLength = Math.min(length, minLength);
sum = sum - nums[leftIndex];
leftIndex++;
length--;
}

}
if (minLength == Integer.MAX_VALUE) {
return 0;
}
return minLength;

}

滑动窗口实现。

代码实现二

public int minSubArrayLen(int s, int[] nums) {

if (nums == null || nums.length == 0) {
return 0;
}
int minLenget = Integer.MAX_VALUE;
int[] sums = new int[nums.length + 1];
sums[0] = 0;
for (int i = 1; i < sums.length; i++) {
sums[i] = sums[i - 1] + nums[i - 1];
}

for (int i = 0; i < sums.length; i++) {
int sum = sums[i] + s;
int nexNum = nextIndex(sums, sum, i);
if (nexNum == sums.length) {
break;
}
minLenget = Math.min(minLenget, nexNum - i);

}
if (minLenget == Integer.MAX_VALUE) {
return 0;
}
return minLenget;
}

/**
* 找出sums中第一个等于大于num的值的下标
*
* @param sums
* @param num
* @return
*/
protected int nextIndex(int[] sums, int num, int left) {

int start = left;
int end = sums.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (sums[mid] == num) {
return mid;
} else if (sums[mid] > num) {

end = mid - 1;

} else if (sums[mid] < num) {
start = mid + 1;
}
}
return start;
}