方法如下：
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1、暴力解法
使用for循环直接逐个求解，算法复杂度为O ( n ) O(n)O(n)
/**
* <p>暴力解法</p>
* @param startInclusive
* @param endExclusive
* @return
*/
public int sumByDirect(int startInclusive, int endExclusive){
int sum = 0;
for (int i = startInclusive; i <p>2、流式编程</p><p>同暴力解法一样，不过使用了声明式的流式编程，代码量更少并且更加的具有可读性</p><pre class="brush:php;toolbar:false"> /**
* <p>流式编程</p><div class="aritcle_card flexRow artxards">
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* @param startInclusive
* @param endExclusive
* @return
*/
public int sumByStream(int startInclusive, int endExclusive){
return IntStream.range(startInclusive, endExclusive).sum();
}3、利用求和公式
利用等差数列求和公式
复杂度为O ( 1 ) O(1)O(1)
/**
* <p>利用求和公式</p>
* @param startInclusive
* @param endExclusive
* @return
*/
public int sumByFormula(int startInclusive, int endExclusive){
return ((startInclusive + endExclusive - 1) * (endExclusive - startInclusive) ) >> 1;
}测试：
@Test
public void Test() {
System.out.println("sumByDirect=" + sumByDirect(1, 101));
System.out.println("sumByStream=" + sumByStream(1, 101));
System.out.println("sumByFormula=" + sumByFormula(1, 101));
}输出结果：
sumByDirect=5050 sumByStream=5050 sumByFormula=5050
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