设 T是n个不等的数构成的数组，现在用分治算法找T的最大数. 先把T从中间划分成两个大小差不多的子数组<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/3078ba9c1b9340f892a68520561faace.png?x-oss-process=image/sharpen,200"/>和<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/adac4f43b43d415c8fd5af1c1f506fcc.png?x-oss-process=image/sharpen,200"/>，递归地求<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/5d788b16e9e7403eaa8fba2c4219d5f6.png?x-oss-process=image/sharpen,200"/>和<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/78f8abde2fd240bf9d6184e5b5e4f4a2.png?x-oss-process=image/sharpen,200"/>的最大数，分别记作<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/cc079551e2d84167b642b57e792e4c39.png?x-oss-process=image/sharpen,200"/>和<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/da86c160b50e45ba9ab1d5129db52714.png?x-oss-process=image/sharpen,200"/>. 比较<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/573632e2a2674a159c22ab2393503d7e.png?x-oss-process=image/sharpen,200"/>和<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/63de55ebddd34ca6ad9e9154868c32aa.png?x-oss-process=image/sharpen,200"/>，那么输出就是<img src="http://jgxm-question.oss-cn-beijing.aliyuncs.com/psst/files/9e00a4c1fa0240609150efd6c3bebca7.png?x-oss-process=image/sharpen,200"/>. <br/> 以元素比较做基本运算，该算法在最坏情况下的时间复杂度的的递推方程是：